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Practice Exam 1

42 multiple-choice and 4 free-response questions, in the format of the AP Precalculus Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time	Calculator
Section I Part A: Multiple Choice (calculator not permitted)	29	65 minutes	Calculator not permitted
Section I Part B: Multiple Choice (graphing calculator required)	13	40 minutes	Graphing calculator required
Section II Part A: Free Response (graphing calculator required)	2	35 minutes	Graphing calculator required
Section II Part B: Free Response (calculator not permitted)	2	35 minutes	Calculator not permitted

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Section I Part A: Multiple Choice (calculator not permitted)

29 questions · 65 minutes · Calculator not permitted

No calculator. The function $g$ is defined for $- 4 \leq x \leq 5$ . Its graph is made of straight segments joining the points $(- 4, 3)$ , $(- 1, 0)$ , $(0, 3)$ , $(3, 1)$ and $(5, 3)$ , in that order. What is the preimage of 3 under $g$ ?
Answer: B.
1. A
  That keeps only the input where the graph meets the $y$ -axis at height 3, as if each output came from one input. A preimage collects every such input, and the graph also reaches 3 at $x = - 4$ and $x = 5$ .
2. B
  Correct. The preimage of 3 is the set of every input whose output is 3. The graph is at height 3 at $x = - 4$ , $x = 0$ and $x = 5$ , and each segment between those points stays below 3.
3. C
  That is the image of 3, $g (3) = 1$ , which runs from input to output. The preimage runs the other way: the inputs that give the output 3.
4. D
  Those are the endpoints, where the graph starts and ends at height 3. It climbs back to 3 at $x = 0$ as well, so 0 belongs in the preimage too.
No calculator. A cyclist rolls past a starting line and keeps speeding up. The table gives her distance past the line, $D (t)$ meters, $t$ seconds after she crosses it.
$\begin{matrix} t & 9 & 9.9 & 10 & 10.1 & 10.5 \\ D (t) & 59.4 & 68.904 & 70 & 71.104 & 75.6 \end{matrix}$
Using the table, which is the best estimate of the rate of change of $D$ at $t = 10$ ?
Answer: C.
1. A
  That is the average rate from $t = 10$ to $t = 10.5$ , $\frac{5.6}{0.5}$ . The interval is longer and lies entirely after $t = 10$ , while she keeps speeding up, so the estimate runs high.
2. B
  That is the average rate from $t = 9$ to $t = 10$ , $\frac{10.6}{1}$ . It covers a whole second before $t = 10$ , when she was slower, so the estimate runs low.
3. C
  Correct. The shortest interval in the table that contains $t = 10$ is from 9.9 to 10.1, and $\frac{71.104 - 68.904}{10.1 - 9.9} = \frac{2.2}{0.2} = 11$ . Average rates over small intervals around a point give the best estimate of the rate at that point.
4. D
  That is $\frac{D (10)}{10}$ , distance divided by time, which treats her speed as if it had been the same since she crossed the line. A rate at a point comes from the change over a small interval around that point.
No calculator. A drone climbs after takeoff. The table gives its altitude, $A (t)$ meters, $t$ seconds after takeoff.
$\begin{matrix} t & 0 & 2 & 4 & 6 & 8 \\ A (t) & 0 & 42 & 76 & 102 & 120 \end{matrix}$
Which statement about the drone's climb from $t = 0$ to $t = 8$ is supported by the table, with a correct reason?
Answer: D.
1. A
  Positive average rates of change say the drone is rising, not how its speed changes. The rates are 21, 17, 13 and 9 meters per second, each smaller than the last, so the climb slows.
2. B
  The gains, 42, 34, 26 and 18 meters over the 2-second intervals, shrink by the same 8 each time, so the rate of change itself is changing. A constant rate would need equal gains every 2 seconds.
3. C
  The conclusion is right, but the reason does not show it. Rising outputs only say the drone gains altitude, and a drone climbing faster and faster would have rising outputs too. What shows the slowing is the average rates of change, which fall from 21 to 9.
4. D
  Correct. Over the four 2-second intervals the average rates of change are $\frac{42 - 0}{2} = 21$ , then 17, 13 and 9 meters per second. They are positive and decreasing, so the drone keeps rising but more slowly, and the graph of $A$ is concave down.
No calculator. A polynomial function $p$ has real coefficients, degree 4 and leading coefficient 3. It has a zero of multiplicity 2 at $x = - 1$ , and $2 i$ is also a zero of $p$ . Which of these is $p (x)$ ?
Answer: B.
1. A
  A zero at $x = - 1$ comes from the factor $x - (- 1) = x + 1$ . The factor $(x - 1)^{2}$ puts the double zero at $x = 1$ instead.
2. B
  Correct. With real coefficients, $- 2 i$ is a zero along with $2 i$ , and $(x - 2 i) (x + 2 i) = x^{2} + 4$ . Together with $(x + 1)^{2}$ for the double zero, that is all four zeros, and the factor 3 sets the leading coefficient.
3. C
  $(x - 2 i) (x + 2 i) = x^{2} - (2 i)^{2} = x^{2} + 4$ , because $i^{2} = - 1$ . The factor $x^{2} - 4$ has the real zeros 2 and $- 2$ , not $2 i$ and $- 2 i$ .
4. D
  Squaring $x - 2 i$ repeats $2 i$ instead of pairing it with its conjugate. Expanded, $(x - 2 i)^{2} = x^{2} - 4 i x - 4$ has a non-real coefficient, and a polynomial with real coefficients must have $- 2 i$ as a zero as well.
No calculator. Let $r (x) = \frac{x^{3} - 2}{4 - 2 x}$ . Which statement describes the end behavior of $r$ ?
Answer: C.
1. A
  That treats $r$ like the cubic on top with its sign flipped. Dividing by the linear denominator lowers the power by one: the leading terms give $- \frac{x^{2}}{2}$ , an even power, so both ends go the same way.
2. B
  A horizontal asymptote at the ratio of the leading coefficients needs the numerator and denominator to have the same degree. Here the numerator's degree is higher, so $r$ is unbounded at both ends.
3. C
  Correct. The numerator has the greater degree, and the quotient of the leading terms is $\frac{x^{3}}{- 2 x} = - \frac{x^{2}}{2}$ . That is negative for inputs of large magnitude of either sign, so $r$ decreases without bound at both ends.
4. D
  That loses a negative sign. The leading term of the denominator is $- 2 x$ , not $2 x$ , so the quotient of the leading terms is $- \frac{x^{2}}{2}$ , which is negative.
No calculator. Which set of values solves $\frac{(x - 2)^{2} (x + 4)}{x - 6} \leq 0$ ?
Answer: D.
1. A
  At $x = 6$ the denominator is 0, so the expression is not defined there. The vertical asymptote is an endpoint of the interval but never a solution.
2. B
  That treats $x = 2$ as a place where the sign changes. The factor $(x - 2)^{2}$ has even multiplicity, so the sign is the same on both sides of 2. For example, $x = 0$ gives $\frac{4 \cdot 4}{- 6} < 0$ , a solution.
3. C
  That leaves out an input where the expression equals 0. The inequality is $\leq 0$ , so the zero at $x = - 4$ is a solution too.
4. D
  Correct. The factor $(x - 2)^{2}$ is never negative, so the sign comes from $\frac{x + 4}{x - 6}$ , which is negative from $- 4$ to 6. The zeros $x = - 4$ and $x = 2$ make the expression 0, so they count, and $x = 6$ is not in the domain.
No calculator. Let $r (x) = \frac{x + 1}{(x - 2)^{2} (x + 3)}$ . Which statement describes $r$ near $x = 2$ ?
Answer: D.
1. A
  That expects a sign change, which happens at a zero of odd multiplicity in the denominator. Here the factor is squared, so it is positive on both sides of 2.
2. B
  Every factor is positive near $x = 2$ : the numerator is near 3, $x + 3$ is near 5, and $(x - 2)^{2}$ is a square. The outputs grow large and positive, not negative.
3. C
  That treats $x = 2$ as a hole. The factor $(x - 2)^{2}$ does not cancel with anything in the numerator, so the denominator alone goes to 0 and the graph has a vertical asymptote there.
4. D
  Correct. Near $x = 2$ the numerator is close to 3 and $x + 3$ is close to 5, while $(x - 2)^{2}$ is a small positive number on both sides. So the outputs are positive and increase without bound from the left and from the right.
No calculator. Let $r (x) = \frac{x^{2} - 4 x}{x^{2} - 16}$ . Which statement about the graph of $r$ at $x = 4$ is correct, with a correct reason?
Answer: A.
1. A
  Correct. $r (x) = \frac{x (x - 4)}{(x - 4) (x + 4)}$ , and the factor $x - 4$ appears once on top and once below, so the graph has a hole at $x = 4$ . For inputs near 4, the outputs are close to $\frac{4}{4 + 4} = \frac{1}{2}$ .
2. B
  The denominator is 0 at $x = 4$ , but so is the numerator, with the same multiplicity. A shared factor with at least the same multiplicity on top gives a hole, not an asymptote. The asymptote is at $x = - 4$ .
3. C
  There is a hole at $x = 4$ , but its height is the value the outputs approach, $\frac{4}{4 + 4} = \frac{1}{2}$ . The 0 comes from the canceled factor and says nothing about that height.
4. D
  $r (4)$ is not defined, so 4 cannot be a zero of $r$ . The real zeros of a rational function are the zeros of the numerator that are in its domain, and the only one here is $x = 0$ .
No calculator. Polynomial long division of $f (x) = 2 x^{3} + x^{2} - 5 x + 4$ by $g (x) = x^{2} + x - 1$ gives $f (x) = g (x) q (x) + r (x)$ , where the degree of $r$ is less than the degree of $g$ . What are $q (x)$ and $r (x)$ ?
Answer: B.
1. A
  In the first subtraction, $- 5 x - (- 2 x) = - 3 x$ , not $- 7 x$ . Subtracting a negative term adds it back; carrying $- 7 x$ forward ends at $- 6 x$ .
2. B
  Correct. $2 x^{3} \div x^{2} = 2 x$ , and subtracting $2 x (x^{2} + x - 1)$ leaves $- x^{2} - 3 x + 4$ . Then $- x^{2} \div x^{2} = - 1$ , and subtracting $- 1 (x^{2} + x - 1)$ leaves $- 2 x + 3$ , whose degree is less than 2, so the division stops.
3. C
  The equation balances, but the division is not finished. This $r$ has degree 2, the same as $g$ , so there is one more step: $- x^{2} \div x^{2} = - 1$ .
4. D
  The last step subtracts $- x^{2} - x + 1$ , and that changes the sign of every term, including the 1: the constant is $4 - 1 = 3$ . Adding the 1 gives 5.
No calculator. The tables give some values of the functions $f$ and $g$ .
$\begin{matrix} x & - 4 & - 2 & 0 & 2 & 4 \\ f (x) & 1 & 5 & 2 & - 1 & 4 \end{matrix}$ $\begin{matrix} x & - 2 & - 1 & 0 & 1 & 2 \\ g (x) & 2 & - 2 & 1 & 4 & - 1 \end{matrix}$
Which expression for $g (x)$ agrees with every value in the tables?
Answer: D.
1. A
  This gives $f (- 4) - 3 = - 2$ at $x = - 2$ , but the table has 2. Every value comes out the negative of the table's, so the reflection over the $x$ -axis is missing.
2. B
  This gives $3 - f (4) = - 1$ at $x = - 2$ , but the table has 2. The negative sign on the input also reflects the graph over the $y$ -axis, which these values do not show.
3. C
  At $x = - 2$ this needs $f (- 1)$ , which is not in the table, and it matches only at $x = 0$ . Multiplying the input by $\frac{1}{2}$ stretches the graph horizontally, but these inputs are half of the ones in the table for $f$ , which is a compression: the input is doubled.
4. D
  Correct. Check each input: $3 - f (- 4) = 2$ , $3 - f (- 2) = - 2$ , $3 - f (0) = 1$ , $3 - f (2) = 4$ and $3 - f (4) = - 1$ . The graph of $f$ is dilated horizontally by a factor of $\frac{1}{2}$ , reflected over the $x$ -axis and translated up 3.
No calculator. An arithmetic sequence $a_{n}$ and a geometric sequence $g_{n}$ with positive terms both have $a_{0} = g_{0} = 4$ and $a_{2} = g_{2} = 36$ . Which statement compares them correctly?
Answer: A.
1. A
  Correct. The common difference is $\frac{36 - 4}{2} = 16$ , so $a_{1} = 20$ and $a_{3} = 52$ . The common ratio has $r^{2} = \frac{36}{4} = 9$ , so $r = 3$ , $g_{1} = 12$ and $g_{3} = 108$ . The geometric sequence starts slower and then pulls ahead, gaining more at each step.
2. B
  Sharing two terms does not make the sequences share the rest. Adding 16 twice and multiplying by 3 twice both take 4 to 36, but they pass through 20 and 12 on the way.
3. C
  The middle terms are $a_{1} = 4 + 16 = 20$ and $g_{1} = 4 \cdot 3 = 12$ , so $a_{1}$ is the larger. The geometric sequence overtakes only after $n = 2$ .
4. D
  $a_{1} = 20$ is larger than $g_{1} = 12$ , but the lead does not last: $a_{3} = 52$ while $g_{3} = 4 \cdot 3^{3} = 108$ . Multiplying by 3 adds more at each step than adding 16 does, once the terms are past 8.
No calculator. After a cleanup of a pond begins, two models are proposed for the mass of a pollutant left in the pond, in kilograms, $t$ weeks later: $L (t) = 800 - 120 t$ and $E (t) = 800 (0.75)^{t}$ . Over which one-week interval does model $E$ first predict a smaller drop in the mass than model $L$ does?
Answer: B.
1. A
  That compares the masses, not the drops. $E (1) = 600$ is already below $L (1) = 680$ , but in that week $E$ drops by 200 kilograms and $L$ by only 120.
2. B
  Correct. Model $L$ drops by 120 kilograms every week. Model $E$ keeps 75% of the mass each week, so it drops by 25% of what is left: $800 - 600 = 200$ , then $600 - 450 = 150$ , then $450 - 337.5 = 112.5$ . The week from $t = 2$ to $t = 3$ is the first with a drop under 120.
3. C
  That is where the masses cross: $E (5) \approx 189.8$ is below $L (5) = 200$ , and $E (6) \approx 142.4$ is above $L (6) = 80$ . The question compares the weekly drops, and $E$ 's drop is already under 120 kilograms in the week from $t = 2$ to $t = 3$ .
4. D
  That takes 25% of the starting 800 kilograms every week, which is how a linear model changes. Model $E$ multiplies the mass by 0.75 each week, so its drops shrink: 200, then 150, then 112.5.
No calculator. Let $g (x) = 5 - 3 \cdot 2^{x}$ . Which statement describes the end behavior of $g$ ?
Answer: A.
1. A
  Correct. As $x$ decreases without bound, $2^{x}$ gets arbitrarily close to 0, so $g (x)$ gets close to 5. As $x$ increases without bound, $3 \cdot 2^{x}$ grows without bound and is subtracted from 5, so $g (x)$ decreases without bound.
2. B
  That swaps the ends. $2^{x}$ is near 0 for large negative $x$ , not large positive $x$ , so the outputs settle near 5 on the left.
3. C
  0 is where $3 \cdot 2^{x}$ heads on the left, but $g$ subtracts that from 5. The vertical translation moves the horizontal asymptote up to $y = 5$ .
4. D
  2 is $g (0) = 5 - 3$ , the $y$ -intercept. To the left of 0 the outputs keep rising toward 5 as $2^{x}$ shrinks toward 0.
No calculator. Which expression is equivalent to $\frac{8^{x}}{2^{x + 3}}$ for every real number $x$ ?
Answer: B.
1. A
  That divides the bases, $\frac{8}{2} = 4$ , and keeps the $- 3$ in the exponent. But $4^{x - 3} = \frac{1}{64} \cdot 4^{x}$ , while the $2^{3}$ in the denominator is only 8.
2. B
  Correct. $8^{x} = (2^{3})^{x} = 2^{3 x}$ , so the quotient is $2^{3 x - (x + 3)} = 2^{2 x - 3} = \frac{(2^{2})^{x}}{2^{3}} = \frac{1}{8} \cdot 4^{x}$ .
3. C
  The 3 in $2^{x + 3}$ sits in the denominator, so $2^{3} = 8$ divides rather than multiplies: the quotient is $\frac{2^{3 x}}{2^{x} \cdot 2^{3}}$ .
4. D
  That adds the exponents $3 x$ and $x$ . Dividing powers of the same base subtracts the exponents: $\frac{2^{3 x}}{2^{x}} = 2^{2 x} = 4^{x}$ .
No calculator. The functions $f$ and $g$ are each defined only for $x = 1, 2, 3, 4, 5$ , and the table gives all of their values.
$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 \\ f (x) & 3 & 5 & 2 & 1 & 4 \\ g (x) & 2 & 4 & 1 & 5 & 3 \end{matrix}$
Let $h (x) = g (f (x))$ . Which list gives $h (1)$ , $h (2)$ , $h (3)$ , $h (4)$ and $h (5)$ , in that order?
Answer: C.
1. A
  That is $f (g (x))$ , with $g$ applied first: $f (g (1)) = f (2) = 5$ . In $g (f (x))$ the inner function is $f$ .
2. B
  That multiplies $f (x)$ by $g (x)$ . A composition feeds each output of $f$ into $g$ as an input; it does not multiply the two functions.
3. C
  Correct. Apply $f$ first, then $g$ : $h (1) = g (3) = 1$ , $h (2) = g (5) = 3$ , $h (3) = g (2) = 4$ , $h (4) = g (1) = 2$ and $h (5) = g (4) = 5$ .
4. D
  That finds each $f (x)$ in the $g (x)$ row and reads the input above it, which runs $g$ backward. For $h (1)$ , $f (1) = 3$ goes into $g$ as an input, and $g (3) = 1$ .
No calculator. The function $f$ is increasing, and the table gives some of its values.
$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 \\ f (x) & - 3 & 0 & 4 & 11 & 20 \end{matrix}$
What value of $x$ solves $f^{- 1} (2 x - 1) = 4$ ?
Answer: A.
1. A
  Correct. $f^{- 1} (2 x - 1) = 4$ means $f (4) = 2 x - 1$ . The table gives $f (4) = 11$ , so $2 x - 1 = 11$ and $x = 6$ .
2. B
  That uses $f^{- 1} (4) = 3$ and solves $2 x - 1 = 3$ . The 4 is an output of $f^{- 1}$ , so it is an input of $f$ , and $2 x - 1 = f (4)$ .
3. C
  That solves $2 x = 11$ and drops the $- 1$ . From $2 x - 1 = 11$ , add 1 first to get $2 x = 12$ .
4. D
  That is $2 f (4) - 1$ , which puts 4 in for $x$ instead of solving for $x$ . The equation says $2 x - 1$ is the input that $f^{- 1}$ sends to 4.
No calculator. A function $f$ is defined for all real numbers, and its graph has no breaks. It increases for $x \leq 1$ , decreases for $1 \leq x \leq 3$ and increases for $x \geq 3$ , with $f (1) = 5$ and $f (3) = - 2$ . Which of the following restrictions of the domain of $f$ makes it invertible, and what is the domain of $f^{- 1}$ for that restriction?
Answer: A.
1. A
  Correct. On $1 \leq x \leq 3$ , $f$ only decreases, so each output comes from exactly one input. Its outputs there run from $f (3) = - 2$ up to $f (1) = 5$ , and that range is the domain of $f^{- 1}$ .
2. B
  $1 \leq x \leq 3$ is a good restriction, but $f^{- 1}$ takes the outputs of $f$ as its inputs. Its domain is the range of $f$ there, from $- 2$ to 5.
3. C
  On $x \geq 1$ , $f$ falls to $- 2$ and then rises again, so an output just above $- 2$ comes from two inputs, one on each side of 3. $f$ is not invertible there.
4. D
  On $0 \leq x \leq 3$ , $f$ rises to 5 at $x = 1$ and then falls, so outputs just below 5 come from two inputs. A domain with $x = 1$ strictly inside it cannot make $f$ invertible.
No calculator. Which of these is equal to $2 \log_{6} 3 + \log_{6} 4$ ?
Answer: D.
1. A
  That turns $2 \log_{6} 3$ into $\log_{6} 6$ by multiplying 3 by 2. The power property raises 3 to the power 2, giving $\log_{6} 9$ .
2. B
  That adds the inputs, $9 + 4$ . A sum of logarithms with the same base is the logarithm of a product: $\log_{6} (9 \cdot 4)$ .
3. C
  That spreads the 2 over both terms, as if the expression were $2 (\log_{6} 3 + \log_{6} 4)$ . The 2 multiplies only $\log_{6} 3$ .
4. D
  Correct. The power property gives $2 \log_{6} 3 = \log_{6} 3^{2} = \log_{6} 9$ . The product property gives $\log_{6} 9 + \log_{6} 4 = \log_{6} 36$ , and $6^{2} = 36$ , so the value is 2.
No calculator. What are all the solutions of $\log_{3} x = \log_{9} (x + 6)$ ?
Answer: C.
1. A
  Both solve $x^{2} = x + 6$ , but $x = - 2$ puts a negative input into $\log_{3} x$ , where it is not defined, even though $\log_{9} (- 2 + 6) = \log_{9} 4$ is defined. Check each candidate in the original equation.
2. B
  That reads $\frac{1}{2} \log_{3} (x + 6)$ as $\log_{3} \frac{x + 6}{2}$ and solves $x = \frac{x + 6}{2}$ . A coefficient of $\frac{1}{2}$ is a power: $\frac{1}{2} \log_{3} (x + 6) = \log_{3} \sqrt{x + 6}$ . At $x = 6$ the two sides are $\log_{3} 6 \approx 1.631$ and $\log_{9} 12 \approx 1.131$ .
3. C
  Correct. Since $9 = 3^{2}$ , $\log_{9} (x + 6) = \frac{\log_{3} (x + 6)}{\log_{3} 9} = \frac{1}{2} \log_{3} (x + 6)$ . Then $2 \log_{3} x = \log_{3} (x + 6)$ , so $x^{2} = x + 6$ and $(x - 3) (x + 2) = 0$ . At $x = 3$ both sides equal 1. At $x = - 2$ , $\log_{3} (- 2)$ is not defined, so $- 2$ is extraneous.
4. D
  That sets the inputs equal, $x = x + 6$ , which has no solution. Equal logarithms have equal inputs only when the bases match, and here $\log_{9} (x + 6)$ is half of $\log_{3} (x + 6)$ .
No calculator. A shuttle train runs back and forth on a straight track between stations A and B. It goes from A to B at a constant speed in 6 minutes, waits 2 minutes at B, returns to A at the same speed in 6 minutes, waits 2 minutes at A, and repeats. Let $d (t)$ be its distance from A, $t$ minutes after it first leaves A. Which statement about the graph of $d$ is true?
Answer: B.
1. A
  $6 + 2 + 6 = 14$ minutes brings the train back to A, but it still waits 2 minutes there before the pattern starts over. The period is 16 minutes.
2. B
  Correct. One full cycle is out, wait, back and wait: $6 + 2 + 6 + 2 = 16$ minutes, after which the train leaves A again. While it waits, $d$ stays at its greatest value at B and at 0 at A, so the graph is flat there.
3. C
  After 8 minutes the train is waiting at B, not leaving A, so the values do not repeat yet. The pattern starts over only after the return trip and the wait at A, at 16 minutes.
4. D
  16 minutes is right, but the train stops twice in each cycle. While it waits, its distance from A does not change, so the graph has a flat piece at B and another at A.
No calculator. The terminal ray of an angle $θ$ in standard position lies in Quadrant II, along the line $y = - 2 x$ . What is $\sin θ$ ?
Answer: D.
1. A
  That takes the sign from the slope, $- 2$ . The sign of sine comes from the vertical displacement of a point on the ray, and in Quadrant II that is positive.
2. B
  That is $\cos θ$ : the horizontal displacement, $- 1$ , over the distance $\sqrt{5}$ . Sine uses the vertical displacement, 2.
3. C
  2 is the $y$ -coordinate of $(- 1, 2)$ , but that point is not on the unit circle. Sine is the vertical displacement divided by the distance from the origin, $\sqrt{5}$ .
4. D
  Correct. The point $(- 1, 2)$ is on the ray, at a distance of $\sqrt{(- 1)^{2} + 2^{2}} = \sqrt{5}$ from the origin. So $\sin θ = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$ , which is positive, as sine is in Quadrant II.
No calculator. A tree stands 30 meters from a ranger station. Its direction from the station is $\frac{2 π}{3}$ radians, measured clockwise from due north. On a map with the station at the origin, east along the positive $x$ -axis and north along the positive $y$ -axis, what are the coordinates of the tree, in meters?
Answer: C.
1. A
  That uses $\frac{2 π}{3}$ as an angle in standard position, counterclockwise from east. Measured clockwise from north, the direction is $\frac{π}{2} - \frac{2 π}{3} = - \frac{π}{6}$ in standard position.
2. B
  That turns counterclockwise from north, to $\frac{π}{2} + \frac{2 π}{3} = \frac{7 π}{6}$ . A clockwise turn subtracts from $\frac{π}{2}$ .
3. C
  Correct. Turning $\frac{2 π}{3}$ clockwise from north gives the standard-position angle $\frac{π}{2} - \frac{2 π}{3} = - \frac{π}{6}$ . The tree is at $(30 \cos (- \frac{π}{6}), 30 \sin (- \frac{π}{6})) = (15 \sqrt{3}, - 15)$ .
4. D
  The sizes of the coordinates are right, but $- \frac{π}{6}$ points below the $x$ -axis, so $30 \sin (- \frac{π}{6}) = - 15$ . The tree is a little south of due east.
No calculator. A sinusoidal function $h$ has a maximum value of 10 at $x = 0$ , a midline of $y = 4$ and a period of 10. On which interval is the graph of $h$ both increasing and concave up?
Answer: A.
1. A
  Correct. With period 10, the minimum is at $x = 5$ and the next maximum at $x = 10$ , so $h$ increases on $5 < x < 10$ , crossing the midline at $x = 7.5$ . The graph is concave up while it is below the midline, from 2.5 to 7.5, and the two intervals overlap on $5 < x < 7.5$ .
2. B
  $h$ is increasing there, but it is above the midline, bending over toward the maximum at $x = 10$ , so the graph is concave down.
3. C
  The graph is concave up there, below the midline, but $h$ is falling toward its minimum at $x = 5$ .
4. D
  Just after the maximum at $x = 0$ , $h$ is decreasing and its graph is concave down, so neither property holds there.
No calculator. The graph of $g$ is made from the graph of $y = \cos θ$ in two steps: first a horizontal dilation by a factor of 3, then a horizontal translation $\frac{π}{2}$ units to the right. Which is $g (θ)$ ?
Answer: A.
1. A
  Correct. The dilation gives $\cos (\frac{θ}{3})$ . Translating that right $\frac{π}{2}$ replaces $θ$ with $θ - \frac{π}{2}$ , which gives a graph with period $6 π$ and a maximum at $θ = \frac{π}{2}$ .
2. B
  This equals $\cos (\frac{1}{3} (θ - \frac{3 π}{2}))$ , a translation of $\frac{3 π}{2}$ , not $\frac{π}{2}$ . The translation replaces $θ$ itself, so it goes inside the factor $\frac{1}{3}$ .
3. C
  Multiplying the input by 3 compresses the graph to period $\frac{2 π}{3}$ . A dilation by a factor of 3 multiplies the input by $\frac{1}{3}$ .
4. D
  Adding $\frac{π}{2}$ to the input moves the graph to the left. A translation to the right subtracts: $θ - \frac{π}{2}$ .
No calculator. Let $g (θ) = 1 - 2 \tan (θ - \frac{π}{4})$ . Its graph has consecutive vertical asymptotes at $θ = - \frac{π}{4}$ and $θ = \frac{3 π}{4}$ . Between those asymptotes, on exactly which interval is the graph of $g$ concave up?
Answer: B.
1. A
  That is where $\tan (θ - \frac{π}{4})$ itself is concave up. The factor $- 2$ reflects the graph over a horizontal line, which swaps concave up and concave down.
2. B
  Correct. Between the asymptotes, $\tan (θ - \frac{π}{4})$ increases, concave down before its point of inflection at $θ = \frac{π}{4}$ and concave up after it. Multiplying by $- 2$ reflects the graph over a horizontal line, so $g$ is concave up on $- \frac{π}{4} < θ < \frac{π}{4}$ .
3. C
  That keeps the point of inflection at $θ = 0$ , where it is for $y = \tan θ$ , and keeps the concavity of tangent. The phase shift moves the point of inflection to $\frac{π}{4}$ , and the reflection flips the concavity.
4. D
  The graph is concave up there, but not only there. The point of inflection moves with the phase shift, to $θ = \frac{π}{4}$ , halfway between the asymptotes. The graph is concave up all the way from $- \frac{π}{4}$ to $\frac{π}{4}$ .
No calculator. Let $f (θ) = 2 \sin θ + 1$ for $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ . Which is $f^{- 1} (x)$ ?
Answer: D.
1. A
  That swaps sine for arcsine and leaves the other steps in place. The inverse undoes the steps in reverse order: subtract 1, divide by 2, then take the arcsine.
2. B
  The inverse undoes the $+ 1$ by subtracting 1. Check with $f (0) = 1$ : $f^{- 1} (1)$ must be 0, but this gives $\arcsin 1 = \frac{π}{2}$ .
3. C
  That is the reciprocal of $f$ , written in $x$ . An inverse function reverses the input-output pairs; it does not divide 1 by the outputs.
4. D
  Correct. Solve $x = 2 \sin θ + 1$ for $θ$ : $\sin θ = \frac{x - 1}{2}$ , and on $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ that gives $θ = \arcsin (\frac{x - 1}{2})$ , for $- 1 \leq x \leq 3$ .
No calculator. What are all the solutions of $2 \cos^{2} x - \cos x = 1$ with $0 \leq x < 2 π$ ?
Answer: B.
1. A
  That solves $\cos x = - \frac{1}{2}$ and loses the other factor. Setting $\cos x - 1 = 0$ adds $x = 0$ , which is in the interval.
2. B
  Correct. Subtract 1 and factor: $(2 \cos x + 1) (\cos x - 1) = 0$ . So $\cos x = - \frac{1}{2}$ , at $x = \frac{2 π}{3}$ and $x = \frac{4 π}{3}$ , or $\cos x = 1$ , at $x = 0$ .
3. C
  Those solve $\cos x = \frac{1}{2}$ and $\cos x = - 1$ , from the factoring $(2 \cos x - 1) (\cos x + 1)$ . Multiplied out, that is $2 \cos^{2} x + \cos x - 1$ , with the wrong sign on the middle term.
4. D
  $\cos x = - \frac{1}{2}$ has two solutions in the interval. Cosine is negative in Quadrants II and III, so $x = \frac{4 π}{3}$ works too.
No calculator. Which expression is equal to $\cos (x + \frac{π}{3}) + \cos (x - \frac{π}{3})$ for every real number $x$ ?
Answer: C.
1. A
  That splits each cosine over the sum, $\cos x + \cos \frac{π}{3}$ and $\cos x - \cos \frac{π}{3}$ , so the $\frac{π}{3}$ parts cancel. The identity multiplies instead, giving each term a factor of $\cos \frac{π}{3} = \frac{1}{2}$ .
2. B
  That uses a minus sign in both expansions. For a difference, $\cos (x - β) = \cos x \cos β + \sin x \sin β$ , so the sine terms cancel instead of adding up.
3. C
  Correct. The sum and difference identities give $\cos x \cos \frac{π}{3} - \sin x \sin \frac{π}{3}$ and $\cos x \cos \frac{π}{3} + \sin x \sin \frac{π}{3}$ . The sine terms cancel, leaving $2 \cos x \cdot \frac{1}{2} = \cos x$ .
4. D
  That takes $\cos \frac{π}{3}$ to be $\frac{\sqrt{3}}{2}$ , which is $\sin \frac{π}{3}$ . With $\cos \frac{π}{3} = \frac{1}{2}$ , the two cosine terms add to $\cos x$ .
No calculator. A polar function $f$ has $f (θ) > 0$ for every $θ$ . As $θ$ increases from 0 to $2 π$ , the polar graph of $r = f (θ)$ traces a closed curve once, counterclockwise. The curve is farthest from the origin, 3 units, at the rectangular points $(3, 0)$ and $(- 3, 0)$ , and closest, 1 unit, at $(0, 1)$ and $(0, - 1)$ . Which could describe the graph of $r = f (θ)$ on rectangular axes, with $θ$ across and $r$ up, for $0 \leq θ \leq 2 π$ ?
Answer: C.
1. A
  One trip around the curve is not one cycle of $r$ . The ray at $θ = \frac{π}{2}$ points straight up, and since $f (θ) > 0$ the point for that angle is on that ray, at $(0, 1)$ . So $f (\frac{π}{2}) = 1$ , and $r$ reaches its least value twice in one turn.
2. B
  Those are the $x$ -coordinates of the points, $r \cos θ$ . The graph of $r$ against $θ$ shows the distance out along each ray, which is positive here and equals 1 at $θ = \frac{π}{2}$ , not 0.
3. C
  Correct. Because $f (θ) > 0$ , the point for each $θ$ lies on the terminal ray of $θ$ , $f (θ)$ units out. The rays at 0, $\frac{π}{2}$ , $π$ , $\frac{3 π}{2}$ and $2 π$ pass through $(3, 0)$ , $(0, 1)$ , $(- 3, 0)$ , $(0, - 1)$ and $(3, 0)$ , so $r$ is 3, 1, 3, 1 and 3 there: two full cycles of $r$ in one turn.
4. D
  Those are the $y$ -coordinates of the points, $r \sin θ$ . The point $(0, - 1)$ is 1 unit from the origin along the ray at $θ = \frac{3 π}{2}$ , so $r = 1$ there, not $- 1$ , and at $θ = 0$ the point is 3 units out, so $r = 3$ , not 0.

Section I Part B: Multiple Choice (graphing calculator required)

13 questions · 40 minutes · Graphing calculator required

Graphing calculator required. Let $p (x) = x^{4} - 2 x^{3} - 5 x^{2} + 4 x + 6$ . What is the global minimum value of $p$ ?
Answer: A.
1. A
  Correct. $p$ has even degree and a positive leading coefficient, so it has a global minimum, and it must be one of the local minimums. The graph turns upward twice, at $x \approx - 1.220$ where $p \approx - 0.475$ and at $x \approx 2.375$ where $p \approx - 7.679$ . The lower of the two, $- 7.679$ , is the global minimum.
2. B
  This is the local minimum at $x \approx - 1.220$ , the first low point you meet moving left to right. The graph dips lower near $x \approx 2.375$ , so this one is not the global minimum.
3. C
  This is the input where the global minimum happens. The question asks for the minimum value, the output $p (2.375) \approx - 7.679$ .
4. D
  This is the local maximum, at $x \approx 0.345$ , where $p$ switches from increasing to decreasing. A minimum is where $p$ switches from decreasing to increasing.
Graphing calculator required. The table gives the values of a function $f$ at seven inputs.
$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ f (x) & 23.425 & 10.350 & 5.225 & 3.550 & 3.825 & 7.550 & 19.225 \end{matrix}$
What is the least degree of a polynomial function whose graph passes through all seven points in the table?
Answer: C.
1. A
  Seven points with distinct inputs always lie on a polynomial of degree 6 or less, so 6 is only an upper bound. The fourth differences of these values are all 3, so a polynomial of lower degree already passes through every point.
2. B
  The fifth differences are all 0. The degree is the level at which the differences first become constant and nonzero, and that is the fourth: 3, 3 and 3.
3. C
  Correct. With the inputs one apart, the differences of the outputs are $- 13.075$ , $- 5.125$ , $- 1.675$ , $0.275$ , $3.725$ and $11.675$ . The second differences are not constant, the third differences are $- 4.5$ , $- 1.5$ , $1.5$ and $4.5$ , and the fourth differences are 3, 3 and 3. Constant nonzero fourth differences mean a degree 4 polynomial passes through all seven points and none of lower degree does. A quartic regression fits every point exactly.
4. D
  The values fall to one low point and rise again, which looks like a parabola, but no quadratic passes through all seven points: the second differences, 7.95, 3.45, 1.95, 3.45 and 7.95, are not constant. A quadratic regression misses $f (2)$ and $f (6)$ by 1.5.
Questions 3 to 4 refer to the following.
A coffee shop counts the customers inside at the start of each hour. Let $t$ be the number of hours after 6 a.m., so $t = 0$ is 6 a.m. and $t = 8$ is 2 p.m. From $t = 0$ to $t = 8$ the counts are 10, 39, 47, 51, 42, 38, 35, 47 and 70. The manager models the count with a cubic regression $C (t)$ .

Graphing calculator required. Which of the following is the cubic regression for the coffee shop data, with $t$ in hours after 6 a.m.?
Answer: A.
1. A
  Correct. Enter $t = 0, 1, 2, \dots, 8$ and the nine counts, then run a cubic regression. Rounded to three decimal places the coefficients are $0.890$ , $- 10.858$ , $37.437$ and $10.343$ .
2. B
  This comes from entering the inputs as 1 to 9. The first count is at $t = 0$ , which is 6 a.m., so every input is one hour too large and the whole model shifts.
3. C
  This comes from entering clock hours, 6 to 14, as the inputs. That model measures time from midnight, but the question defines $t$ as hours after 6 a.m.
4. D
  This is a quadratic regression. The counts rise, fall and rise again, two turns that a quadratic cannot show, and the question asks for a cubic.
Uses the material above question 3.

Graphing calculator required. Use the cubic regression $C$ as your calculator stores it, without rounding its coefficients. What is the average rate of change of $C$ from 7:30 a.m. to 12:30 p.m.?
Answer: B.
1. A
  That is $C (6.5) - C (1.5) \approx - 5.773$ , the change in the count over the interval, in customers. An average rate of change divides the change by the 5 hours it took.
2. B
  Correct. 7:30 a.m. is $t = 1.5$ and 12:30 p.m. is $t = 6.5$ . The model gives $C (1.5) \approx 45.073$ and $C (6.5) \approx 39.300$ , so the average rate of change is $\frac{C (6.5) - C (1.5)}{6.5 - 1.5} \approx \frac{- 5.773}{5} \approx - 1.155$ customers per hour.
3. C
  That uses the counts at 7 a.m. and noon from the table, $\frac{35 - 39}{5} = - 0.8$ . The question asks about the model at 7:30 and 12:30, times the table does not have, which is what the model is for.
4. D
  That subtracts in the wrong order, $C (1.5) - C (6.5)$ over 5. The model's count is lower at 12:30 than at 7:30, so the average rate of change is negative.
Graphing calculator required. The population of Ashford is modeled by $A (t) = 18400 (1.032)^{t}$ and the population of Brookvale by $B (t) = 25100 (1.011)^{t}$ , where $t$ is the number of years after July 1, 2020. Each model gives the town's population on July 1. According to the models, in which year is Ashford's population on July 1 greater than Brookvale's for the first time?
Answer: C.
1. A
  This solves $18400 (1.032)^{t} = 25100$ , which treats Brookvale as if it stayed at 25,100 people. Brookvale grows too, so Ashford takes longer to catch it: about 15.104 years, not 9.858.
2. B
  The models are equal at $t \approx 15.104$ , a little after July 1, 2035. On July 1, 2035, $t = 15$ , Ashford's model gives about 29,513 people and Brookvale's about 29,576, so Ashford is still smaller that day.
3. C
  Correct. The graphs of $A$ and $B$ intersect at $t \approx 15.104$ . The first July 1 after that is $t = 16$ , in 2036, when $A (16) \approx 30457$ and $B (16) \approx 29901$ .
4. D
  This treats the growth as linear, adding 3.2% and 1.1% of the 2020 populations every year, which makes the towns equal at $t \approx 21.426$ . A percent growth rate multiplies the current population each year, so the models are exponential.
Graphing calculator required. A climbing gym had 6.2, 7.1, 7.6, 8.9, 9.5, 10.6 and 11.4 hundred members at the end of years $t = 0, 1, 2, \dots, 6$ . The owner fits a linear regression and an exponential regression to these data. The error of a model at a data point is the absolute value of its residual. Which model has the smaller largest error over the seven data points, and what is that largest error, in hundreds of members?
Answer: A.
1. A
  Correct. The linear regression is $L (t) \approx 6.132 + 0.875 t$ , and its residuals run from $- 0.282$ at $t = 2$ to $0.143$ at $t = 3$ , so its largest error is 0.282. The exponential regression is $E (t) \approx 6.320 (1.107)^{t}$ , and its residuals run from $- 0.244$ at $t = 6$ to $0.322$ at $t = 3$ , so its largest error is 0.322. The linear model's worst miss is the smaller one.
2. B
  0.143 is the linear model's largest residual as a signed number. The error is the absolute value, and the residual of $- 0.282$ at $t = 2$ is the bigger miss.
3. C
  This compares only the negative residuals, $- 0.282$ for the linear model and $- 0.244$ for the exponential one. The exponential model also misses by 0.322 at $t = 3$ , so its largest error is 0.322.
4. D
  0.322 is the largest error of the calculator's exponential regression, but it is the larger of the two. A gym's membership sounds like percent growth, yet on these seven years the linear model's largest error, 0.282, is smaller.
Graphing calculator required. What are all the solutions of $\log_{2} (x + 5) = 0.4 x$ ?
Answer: D.
1. A
  This misses the solution near the vertical asymptote at $x = - 5$ , where the logarithm's graph climbs steeply and is easy to overlook. Zoom in on $- 5 < x < - 4$ and the graphs meet again at $x \approx - 4.731$ .
2. B
  This uses the base 10 logarithm, the calculator's plain log key, in place of $\log_{2}$ . Enter $\log_{2} (x + 5)$ as $\frac{\ln (x + 5)}{\ln 2}$ or with a base 2 logarithm.
3. C
  This solves $\log_{2} (x) + 5 = 0.4 x$ , with the parenthesis closed too early. The input of the logarithm is $x + 5$ , so its graph starts at $x = - 5$ , not at $x = 0$ .
4. D
  Correct. Graph $y = \frac{\ln (x + 5)}{\ln 2}$ and $y = 0.4 x$ . The logarithm is defined for $x > - 5$ , and its graph crosses the line just to the right of the asymptote, at $x \approx - 4.731$ , and again at $x \approx 9.693$ . Both inputs are in the domain, so both are solutions.
Graphing calculator required. A phone's battery charge, as a percent of full, was recorded $m$ minutes after the phone was plugged in: 17 at $m = 5$ , 36 at $m = 10$ , 47 at $m = 15$ , 52 at $m = 20$ , 64 at $m = 30$ , 72 at $m = 40$ and 80 at $m = 60$ . Which of the following is the logarithmic regression $P (m) = a + b \ln m$ for these data?
Answer: A.
1. A
  Correct. Run a logarithmic regression with the minutes as the input list and the charge as the output list. The calculator reports $a \approx - 23.305$ and $b \approx 25.535$ .
2. B
  This swaps $a$ and $b$ . In $a + b \ln m$ the coefficient $b$ multiplies $\ln m$ , and this version has the charge falling the longer the phone is plugged in.
3. C
  This comes from a linear regression of the charge on $\log m$ , base 10, with the answer then written with $\ln m$ . Changing the base changes $b$ by a factor of $\ln 10$ : $25.535 \ln 10 \approx 58.797$ . Written with $\ln m$ , the coefficient is 25.535.
4. D
  These are the coefficients of a linear regression, $25.633 + 1.048 m$ , put into a logarithmic form. A linear fit and a logarithmic fit give different values of $a$ and $b$ .
Graphing calculator required. A rare trading card sold at auction for 42, 55, 79, 106, 151 and 208 dollars at $t = 0, 2, 4, 6, 8$ and 10 years after it was printed. A collector plots the sales on a semi-log plot whose vertical axis is scaled with the natural logarithm, so a sale for $y$ dollars sits at height $\ln y$ . Which equation is the linear regression line for the points $(t, \ln y)$ on that plot?
Answer: C.
1. A
  This uses base 10 logarithms of the prices. The plot's vertical axis is scaled with the natural logarithm, so the heights are $\ln 42 \approx 3.738$ , $\ln 55 \approx 4.007$ and so on, not $\log 42 \approx 1.623$ .
2. B
  These are the $b$ and $a$ of the exponential model $y \approx 40.990 (1.176)^{t}$ . On the semi-log plot the slope is $\ln b$ and the intercept is $\ln a$ , not $b$ and $a$ themselves.
3. C
  Correct. Store $\ln y$ for each price, 3.738, 4.007, 4.369, 4.663, 5.017 and 5.338, and run a linear regression of those values on $t$ . The slope is about 0.162 and the intercept about 3.713.
4. D
  This swaps the slope and the intercept. The intercept is the line's height at $t = 0$ , which should be near $\ln 42 \approx 3.738$ , so 3.713 is the intercept and 0.162 is the slope.
Graphing calculator required. An astronomer records the brightness of a variable star once a night, in arbitrary units. On nights $x = 0, 1, 2, \dots, 14$ the readings are 4.89, 5.13, 5.16, 4.82, 4.14, 3.53, 2.97, 2.84, 2.92, 3.42, 4.01, 4.66, 5.04, 5.22 and 4.96. According to a sinusoidal regression on these data, what is the period of the star's brightness cycle, in nights?
Answer: D.
1. A
  0.550 is $b$ , the coefficient of $x$ in $y = a \sin (b x + c) + d$ , not the period. The period is $\frac{2 π}{b}$ .
2. B
  That reads the period from the table, from the high reading on night 2 to the one on night 13. The regression uses all 15 readings, and its period is $\frac{2 π}{b}$ , about 11.430.
3. C
  This is $\frac{π}{b}$ , half the period, which is the time from a maximum to the next minimum. The cycle repeats after a full period, from one maximum to the next.
4. D
  Correct. The regression gives $y \approx 1.203 \sin (0.550 x + 0.803) + 4.002$ , though a calculator may show a different but equivalent value of $c$ . With the stored $b \approx 0.54971$ , the period is $\frac{2 π}{b} \approx 11.430$ nights, which is close to the 11 nights between the two highest readings in the table, on nights 2 and 13.
Graphing calculator required. A horse on a carousel is $h (t) = 1.1 + 0.35 \sin (\frac{2 π t}{3.2})$ meters above the platform $t$ seconds after the ride starts. The ride lasts 48.4 seconds. How many times during the ride does the horse pass a height of 1.3 meters while moving up?
Answer: B.
1. A
  This counts only the 15 complete cycles, which take 48 seconds. The ride goes on for 0.4 second into a 16th cycle, and the horse rises through 1.3 meters 0.310 second into each cycle, so it does that once more, at $t \approx 48.310$ , before the ride stops.
2. B
  Correct. Solving $h (t) = 1.3$ gives $\sin (\frac{2 π t}{3.2}) = \frac{0.2}{0.35} \approx 0.571$ , so the first upward pass is at $t = \frac{3.2}{2 π} \arcsin (\frac{0.2}{0.35}) \approx 0.310$ . There is one upward pass every 3.2 seconds after that, at $t \approx 0.310 + 3.2 k$ . The last one before 48.4 seconds is $k = 15$ , at $t \approx 48.310$ , so there are 16.
3. C
  This counts passes in both directions as two per cycle and multiplies by $\frac{48.4}{3.2} = 15.125$ cycles, rounding 30.25 down. The question asks only for passes on the way up, one per cycle.
4. D
  The horse is at 1.3 meters 31 times during the ride, but that counts it coming down as well as going up. Only 16 of the 31 are on the way up.
Graphing calculator required. Points $P$ and $Q$ have polar coordinates $(4.2, 0.6)$ and $(- 2.5, 1.9)$ , with angles in radians. What is the distance between $P$ and $Q$ ?
Answer: B.
1. A
  This places $Q$ at $(2.5, 1.9)$ , on the terminal ray of the angle 1.9. A negative $r$ puts the point on the opposite ray, at about $(0.808, - 2.366)$ in rectangular coordinates.
2. B
  Correct. Convert each point with $x = r \cos θ$ and $y = r \sin θ$ : $P \approx (3.466, 2.371)$ and $Q \approx (0.808, - 2.366)$ . The distance is $\sqrt{(3.466 - 0.808)^{2} + (2.371 + 2.366)^{2}} \approx 5.432$ .
3. C
  This comes from a calculator in degree mode, which reads 0.6 and 1.9 as degrees. Both points then sit almost on the $x$ -axis on opposite sides of the origin, about $4.2 + 2.5$ apart. The angles are in radians.
4. D
  This treats $(r, θ)$ as rectangular coordinates: $\sqrt{(4.2 + 2.5)^{2} + (0.6 - 1.9)^{2}} \approx 6.825$ . Convert to rectangular coordinates before using the distance formula.
Graphing calculator required. The polar function $f$ is given by $f (θ) = 1 + 0.8 θ - 3 \sin (1.3 θ)$ for $0 \leq θ \leq 2$ . What is the greatest distance between the origin and a point $(f (θ), θ)$ on the polar graph of $f$ ?
Answer: D.
1. A
  This takes the low point to be where $\sin (1.3 θ) = 1$ , at $θ = \frac{π}{2.6} \approx 1.208$ , which ignores the $0.8 θ$ term. There $| f | \approx 1.033$ , but $f$ reaches its minimum earlier, at $θ \approx 1.049$ .
2. B
  1.049 is the value of $θ$ where $f$ reaches its minimum. The distance from the origin is the size of the output there, $| f (1.049) | \approx 1.097$ .
3. C
  1.053 is $f (2)$ , the greatest value of $f$ on the interval. Distance from the origin is $| f (θ) |$ , and $f$ falls to about $- 1.097$ , which is farther from the origin than 1.053.
4. D
  Correct. The distance from the origin to $(f (θ), θ)$ is $| f (θ) |$ . On the interval $f$ starts at $f (0) = 1$ , falls to a minimum of about $- 1.097$ at $θ \approx 1.049$ , then rises to $f (2) \approx 1.053$ . The largest size of $f$ is at that minimum, so the greatest distance is about 1.097.

Section II Part A: Free Response (graphing calculator required)

2 questions · 35 minutes · Graphing calculator required

Question 1: Function Concepts 6 points

Let $f$ be an increasing function defined for all real numbers. The table gives values of $f (x)$ at selected values of $x$ . The function $g$ is given by $g (x) = \frac{4 x^{2} - 3 x - 5}{2 x^{2} + 7}$ .

Selected values of $f (x)$
$x$ 0 3 6 9 12

$f (x)$ 64 80 100 125 156.25

A graphing calculator is required for this question. Use it to evaluate $g$ , to find its zeros, and to check its behavior on a graph or a table. Any decimal answer should be accurate to three places after the decimal point. Respond to parts A, B, and C, labeling subparts i and ii.
1. A.
  
  i. Find the value of $(g \circ f) (6)$ as a decimal approximation, or indicate that it is not defined.
  
  ii. Find the value of $f^{- 1} (80)$ , or indicate that it is not defined.
  Question 1, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Finds $(g \circ f) (6) = g (f (6)) = g (100) \approx 1.984$ .Earns it: "f(6) = 100, so g(100) = 1.984"; " $(g \circ f) (6) \approx 1.984$ " with $g (100)$ shownDoes not: " $f (g (6))$ " (the order is reversed: $f$ acts first, then $g$ ); " $g (6) \cdot f (6)$ " (a product, not a composition); "1.98" (not accurate to three decimal places)
  - 1 point Finds $f^{- 1} (80) = 3$ . Because $f$ is increasing, it is one-to-one and $f^{- 1}$ exists, and $f (3) = 80$ .Earns it: " $f (3) = 80$ , so $f^{- 1} (80) = 3$ "; "3"Does not: " $\frac{1}{80}$ " (reads $f^{- 1}$ as a reciprocal); " $\frac{1}{f (80)}$ " or "not defined because 80 is not an input in the table" (80 is an output of $f$ , so it is an input of $f^{- 1}$ )
  Model response
  i. $f (6) = 100$ from the table, so $(g \circ f) (6) = g (100) = \frac{4 (100)^{2} - 3 (100) - 5}{2 (100)^{2} + 7} = \frac{39695}{20007} \approx 1.984$ .
  
  ii. $f$ is increasing, so each output comes from exactly one input and $f^{- 1}$ exists. The table gives $f (3) = 80$ , so $f^{- 1} (80) = 3$ .
2. B.
  
  i. Find all real zeros of $g$ . Express each zero as a decimal approximation.
  
  ii. Determine the end behavior of $g$ as $x$ increases without bound. Express your answer using the mathematical notation of a limit.
  Question 1, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Finds both real zeros, $x \approx - 0.804$ and $x \approx 1.554$ , and no others. The denominator $2 x^{2} + 7$ is never 0, so each zero of the numerator is a zero of $g$ . Exact values $\frac{3 \pm \sqrt{89}}{8}$ also earn the point.Earns it: " $x = - 0.804$ and $x = 1.554$ " from a graph or a solver; " $4 x^{2} - 3 x - 5 = 0$ , so $x = \frac{3 \pm \sqrt{89}}{8}$ , about $- 0.804$ and $1.554$ "Does not: only one of the two zeros; " $x = \pm \sqrt{3.5}$ " (sets the denominator equal to 0, which has no real solutions); " $g (0) = - \frac{5}{7}$ " (that is the vertical intercept, not a zero)
  - 1 point States $\lim_{x \to \infty} g (x) = 2$ , the ratio of the leading coefficients $\frac{4}{2}$ , since the numerator and denominator have the same degree.Earns it: " $\lim_{x \to \infty} g (x) = 2$ "; " $\lim_{x \to \infty} \frac{4 x^{2} - 3 x - 5}{2 x^{2} + 7} = \frac{4}{2} = 2$ "Does not: " $y = 2$ " or "g approaches 2" (correct value but not in limit notation); " $\lim_{x \to \infty} g (x) = \infty$ " (the degrees are equal, so the outputs level off); " $\lim_{x \to - \infty} g (x) = 2$ " (the question asks about $x$ increasing without bound)
  Model response
  i. $2 x^{2} + 7 > 0$ for every $x$ , so $g$ is 0 exactly where $4 x^{2} - 3 x - 5 = 0$ . Graphing $g$ and using the zero feature gives $x \approx - 0.804$ and $x \approx 1.554$ .
  
  ii. The numerator and denominator are both degree 2, so for large $x$ the outputs approach the ratio of the leading coefficients, $\frac{4}{2}$ . $\lim_{x \to \infty} g (x) = 2$ .
3. C.
  
  i. Of the function types linear, quadratic, and exponential, which best models $f$ ?
  
  ii. Give a reason for your answer in part C (i) based on how the output values of $f$ change over the equal-length input-value intervals in the table. Refer to values in the table in your reasoning.
  Question 1, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Identifies an exponential function. The point is for the function type alone; an expression is not asked for and is not scored.Earns it: "Exponential"; "Exponential, $f (x) = 64 (1.25)^{x / 3}$ "Does not: "Quadratic" (the second differences 4, 5 and 6.25 are not constant); "Linear" (the first differences 16, 20, 25 and 31.25 are not constant)
  - 1 point Gives a reason based on constant ratios: over each input interval of length 3, the output values are multiplied by the same factor, $\frac{80}{64} = \frac{100}{80} = \frac{125}{100} = \frac{156.25}{125} = 1.25$ . Outputs that change proportionally over equal-length intervals are what an exponential function produces.Earns it: "Every time $x$ goes up by 3, $f (x)$ is multiplied by 1.25: 80/64, 100/80, 125/100 and 156.25/125 all equal 1.25."Does not: "The first differences 16, 20, 25, 31.25 are not constant, so it is not linear" (rules out one type but does not show the outputs change by a constant factor); "The second differences 4, 5, 6.25 are not constant" alone (rules out quadratic but does not establish exponential); "The values increase faster and faster" (an increasing quadratic could do that too)
  Model response
  i. Exponential.
  
  ii. The inputs go up by 3 each step, and the ratios of consecutive outputs are $\frac{80}{64} = 1.25$ , $\frac{100}{80} = 1.25$ , $\frac{125}{100} = 1.25$ and $\frac{156.25}{125} = 1.25$ . The outputs are multiplied by the same factor over equal-length input intervals, so they change proportionally and an exponential model fits. For example, $f (x) = 64 (1.25)^{x / 3}$ matches all five values.
Your score: 0 of 6, from the boxes you ticked.
Question 2: Modeling a Non-Periodic Context 6 points

A veterinarian records the mass of a puppy as it grows. At 3 months old, at $t = 3$ , the puppy has a mass of 14.8 kilograms. At 8 months old, at $t = 8$ , its mass is 25.0 kilograms. The puppy's mass, in kilograms, $t$ months after it was born can be modeled by the function $m$ given by $m (t) = 30 - b c^{t}$ , where $b$ and $c$ are positive constants and $c < 1$ .

A graphing calculator is required for this question. Use it to solve for the constants, to evaluate $m$ , and to solve equations. Store values instead of rounding along the way, and give decimal answers accurate to three places after the decimal point. Respond to parts A, B, and C, labeling subparts i and ii.
1. A.
  
  i. Use the given data to write two equations that can be used to find the values of the constants $b$ and $c$ in the expression for $m (t)$ .
  
  ii. Find the values of $b$ and $c$ as decimal approximations.
  Question 2, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Writes the two equations $30 - b c^{3} = 14.8$ and $30 - b c^{8} = 25.0$ , or the equivalent pair $b c^{3} = 15.2$ and $b c^{8} = 5$ .Earns it: " $m (3) = 30 - b c^{3} = 14.8$ and $m (8) = 30 - b c^{8} = 25$ "; " $b c^{3} = 15.2$ and $b c^{8} = 5$ "Does not: " $b c^{3} = 14.8$ and $b c^{8} = 25$ " (drops the 30: the model's output is $30 - b c^{t}$ , not $b c^{t}$ ); " $30 - b c^{3} = 14.8$ " alone (two equations are needed)
  - 1 point Finds $c \approx 0.801$ and $b \approx 29.619$ . Dividing $b c^{8} = 5$ by $b c^{3} = 15.2$ gives $c^{5} = \frac{5}{15.2}$ , so $c = {(\frac{5}{15.2})}^{1 / 5}$ and $b = \frac{15.2}{c^{3}}$ . The truncated values $c = 0.800$ and $b = 29.618$ also earn the point. Any method of solving the system earns the point, including a graphing calculator's intersection or solver feature.Earns it: " $c^{5} = \frac{5}{15.2}$ , $c = 0.801$ , $b = \frac{15.2}{c^{3}} = 29.619$ "; " $b = 29.619$ , $c = 0.801$ " from a calculator solverDoes not: " $c = 1.111$ " (that is ${(\frac{25}{14.8})}^{1 / 5}$ , which treats the masses themselves as exponential; the part that changes by a constant factor is $30 - m (t)$ ); " $c = 0.329$ " (that is $\frac{5}{15.2}$ , the factor over 5 months, not over 1 month)
  Model response
  i. $m (3) = 14.8$ and $m (8) = 25.0$ give $30 - b c^{3} = 14.8 and 30 - b c^{8} = 25.0,$ that is, $b c^{3} = 15.2$ and $b c^{8} = 5$ .
  
  ii. Dividing the second equation by the first gives $c^{5} = \frac{5}{15.2}$ , so $c = {(\frac{5}{15.2})}^{1 / 5} \approx 0.801$ . Then $b = \frac{15.2}{c^{3}} \approx 29.619$ . So $m (t) \approx 30 - 29.619 (0.801)^{t}$ , with the unrounded values stored in the calculator.
2. B.
  
  i. Use model $m$ to find the average rate of change of the puppy's mass, in kilograms per month, from $t = 6$ to $t = 12$ months. Express your answer as a decimal approximation. Show the computations that lead to your answer.
  
  ii. Interpret the meaning of your answer from part B (i) in the context of the problem.
  Question 2, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Computes $\frac{m (12) - m (6)}{12 - 6} = \frac{27.946 - 22.200}{6} \approx 0.958$ kilogram per month, with the computation shown. 0.957 also earns the point.Earns it: " $\frac{m (12) - m (6)}{6} = \frac{27.946 - 22.200}{6} = 0.958$ "Does not: " $\frac{25.0 - 14.8}{8 - 3} = 2.04$ " (the average rate over the data, not over 6 to 12 months with the model); "0.959" (comes from rounding $b$ and $c$ to three places before computing, which is not accurate to three places); " $m (12) - m (6) = 5.746$ " (a change, not a rate)
  - 1 point Interprets the value in context with units and the time interval: from 6 months to 12 months old, the puppy's mass increases by about 0.958 kilogram per month, on average.Earns it: "Between 6 and 12 months old, the puppy's mass grew by an average of 0.958 kilogram per month."Does not: "The puppy gains 0.958 kilogram every month" (no time interval, and treats the rate as constant); "The mass increases by 0.958" (no "per month" and no interval)
  Model response
  i. $m (6) \approx 22.200$ and $m (12) \approx 27.946$ , so the average rate of change is $\frac{m (12) - m (6)}{12 - 6} = \frac{27.946 - 22.200}{6} \approx 0.958 kilogram per month.$
  
  ii. From 6 months to 12 months old, the puppy's mass increased by an average of about 0.958 kilogram per month.
3. C.
  
  Use model $m$ to respond to the following.
  
  i. Write and solve an equation whose solution gives the time $t$ , in months, when the model predicts a mass of 28 kilograms. Express your value for $t$ as a decimal approximation.
  
  ii. A neighbor starts from the model's mass at $t = 12$ months and adds the average rate from part B (i), about 0.958 kilogram per month, for each of the 6 months from $t = 12$ to $t = 18$ . The neighbor says that this predicts a mass of more than 32 kilograms at $t = 18$ months. Explain why model $m$ does not support this prediction.
  Question 2, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Writes $30 - b c^{t} = 28$ with the values from part A, and solves it: $t \approx 12.121$ months. 12.120 also earns the point.Earns it: " $30 - 29.619 (0.801)^{t} = 28$ , $t = 12.121$ " with stored values; " $b c^{t} = 2$ , so $t = \frac{\ln (2 / b)}{\ln c} = 12.121$ "Does not: " $t = 12.121$ " with no equation; " $t = 12.147$ " (comes from rounding $b$ and $c$ before solving); " $b c^{t} = 28$ " (sets the changing part equal to 28 instead of the mass)
  - 1 point Explains, using the model, why it never gives a mass above 30 kilograms: $b > 0$ and $c^{t} > 0$ , so $b c^{t} > 0$ and $m (t) < 30$ for every $t$ , and as $t$ increases without bound $m (t)$ approaches 30. So the model cannot reach 32 kilograms at $t = 18$ or at any time. An explanation that computes $m (18) \approx 29.459$ , compares it with 32, and says that the model's monthly gains shrink, so the average rate from part B (i) does not continue, also earns the point. The neighbor's projection itself is $m (12) + 6 r \approx 33.692$ kilograms, where $r \approx 0.958$ is the stored rate from part B (i); computing it is not required.Earns it: " $b c^{t}$ is always positive, so $m (t)$ is always less than 30. The model never reaches 32 kilograms."; " $\lim_{t \to \infty} m (t) = 30$ , and $m$ rises toward 30 without passing it, so the model never predicts more than 30 kilograms."; "The model gives $m (18) \approx 29.459$ , less than 32. Its gains get smaller each month, so continuing at 0.958 kilogram per month overstates the growth."Does not: "32 kilograms is too heavy for a dog" (does not use the model); "The rate from part B (i) is only an average" alone (does not say what the model predicts); " $m (18) = 29.459$ " with no comparison with 32 and no reason
  Model response
  i. $30 - b c^{t} = 28$ , using the stored values of $b$ and $c$ , so $b c^{t} = 2$ . Graphing $y = m (t)$ and $y = 28$ and finding the intersection, or solving $t = \frac{\ln (2 / b)}{\ln c}$ , gives $t \approx 12.121$ months.
  
  ii. Because $b > 0$ and $c^{t} > 0$ , the term $b c^{t}$ is positive for every $t$ , so $m (t) = 30 - b c^{t}$ is less than 30 for every $t$ . As $t$ increases, $b c^{t}$ shrinks toward 0, so the model's mass rises toward 30 kilograms and gains less each month. At $t = 18$ the model gives $m (18) \approx 29.459$ kilograms. The neighbor's projection, $m (12) + 6 r \approx 33.692$ kilograms with $r \approx 0.958$ the rate from part B (i), assumes the gains stay at 0.958 kilogram per month, but the model's gains keep shrinking. Model $m$ never predicts more than 30 kilograms, so it does not support a mass above 32 kilograms.
Your score: 0 of 6, from the boxes you ticked.

Selected values of $f (x)$
$x$	0	3	6	9	12
$f (x)$	64	80	100	125	156.25

Your results

Section	Score	Time used
Section I Part A
Section I Part B
Section II Part A
Section II Part B

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Practice Exam 2

42 multiple-choice and 4 free-response questions, in the format of the AP Precalculus Exam. Each section runs on its own clock and ends when time runs out.

Section	Questions	Time	Calculator
Section I Part A: Multiple Choice (calculator not permitted)	29	65 minutes	Calculator not permitted
Section I Part B: Multiple Choice (graphing calculator required)	13	40 minutes	Graphing calculator required
Section II Part A: Free Response (graphing calculator required)	2	35 minutes	Graphing calculator required
Section II Part B: Free Response (calculator not permitted)	2	35 minutes	Calculator not permitted

One section at a time. When a section's clock reaches zero, the section ends and is scored.
Multiple choice is scored when its section ends, with an explanation for every choice.
Free response is never scored by the page. When the section ends, you see the scoring guide and a model response for each part and tick the points your answer earns.
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Time for each section Standard time Time and a half Double time Untimed Study mode

Section I Part A: Multiple Choice (calculator not permitted)

29 questions · 65 minutes · Calculator not permitted

No calculator. Maya walks at a constant speed along a straight sidewalk. A lamppost stands 6 meters from the sidewalk, and partway through her walk she passes the point of the sidewalk closest to the post. Let $D (t)$ be her distance from the lamppost, in meters, $t$ seconds after she starts. Which describes the graph of $D$ over her walk?
Answer: D.
1. A
  That is the graph of her distance from the closest point of the sidewalk, which does reach 0. The lamppost is 6 meters off the sidewalk, so she is never closer to it than 6 meters.
2. B
  A graph that is concave down has a rate of change that keeps falling. Here the rate of change starts negative, passes 0 beside the post and ends positive, so it rises, which makes the graph concave up.
3. C
  Straight pieces would mean her distance from the post changes at a constant rate. Near the closest point her steps run almost sideways to the post, so the distance barely changes there, while far away it changes at nearly her full speed.
4. D
  Correct. Far from the post nearly all of her motion is toward it or away from it, so $D$ falls at close to her walking speed at the start and rises at close to it at the end. Beside the post her steps are almost sideways to it, so $D$ changes slowly and reaches its least value, 6. The rate of change of $D$ climbs the whole time, from about minus her speed to about plus her speed, so the graph is concave up throughout.
No calculator. In a city, the number of people who ride the bus each day falls as the price of a bus ticket rises. The number of cars on the city's roads each day rises as the number of bus riders falls. Treat the number of cars as a function of the ticket price. What is the sign of its rate of change?
Answer: A.
1. A
  Correct. A higher price means fewer riders, and fewer riders means more cars. So as the price increases, the number of cars increases too, and when two quantities increase together the rate of change of one with respect to the other is positive.
2. B
  That is the sign of the rate of change of riders with respect to price. The question asks about cars, and cars move the opposite way from riders.
3. C
  That is the sign of the rate of change of cars with respect to riders. Follow the chain one more step: price up, riders down, cars up, so cars rise with price.
4. D
  The sign of a rate of change comes from how the quantities move together, not from whether their values are positive. A positive quantity can have a negative rate of change.
No calculator. A polynomial function $p$ switches between increasing and decreasing at exactly two points: it has a local maximum at $(- 1, 6)$ and a local minimum at $(2, - 21)$ . Also, $\lim_{x \to \infty} p (x) = \infty$ . How many real zeros does $p$ have?
Answer: C.
1. A
  That counts only the zero between the two turning points, where the graph falls from 6 to $- 21$ . The graph also crosses the $x$ -axis once to the left of $x = - 1$ and once to the right of $x = 2$ .
2. B
  That counts the zero between the turning points and the one after the minimum, where the graph rises from $- 21$ without bound. To the left of $x = - 1$ the graph is increasing up to 6, so it comes up from below the $x$ -axis and crosses it there too.
3. C
  Correct. With no other turning points, $p$ increases for $x < - 1$ , decreases from $x = - 1$ to $x = 2$ , and increases for $x > 2$ . Increasing up to the value 6 at $x = - 1$ with no turn before it means $\lim_{x \to - \infty} p (x) = - \infty$ . Each of the three pieces is always increasing or always decreasing and passes through 0 once: from below 0 up to 6, from 6 down to $- 21$ , and from $- 21$ up without bound.
4. D
  Four zeros would need the graph to come down from above on the far left, as an even-degree graph with both ends up does. Then $p$ would have to turn again somewhere left of $x = - 1$ , but it has only the two turning points given.
No calculator. Which of the following is $p (x) = x^{4} + 5 x^{2} - 36$ written as a product of linear factors?
Answer: C.
1. A
  That factors $x^{2} + 9$ as if it were $x^{2} - 9$ . The product here is $(x^{2} - 4) (x^{2} - 9) = x^{4} - 13 x^{2} + 36$ , not $p (x)$ .
2. B
  The factor pair 36 and 1 does multiply to 36, but $1 - 36 = - 35$ , not 5. This product is $(x^{2} - 36) (x^{2} + 1) = x^{4} - 35 x^{2} - 36$ .
3. C
  Correct. As a quadratic in $x^{2}$ , $x^{4} + 5 x^{2} - 36 = (x^{2} + 9) (x^{2} - 4)$ , since $9 \cdot (- 4) = - 36$ and $9 + (- 4) = 5$ . Then $x^{2} - 4 = (x - 2) (x + 2)$ , and $x^{2} + 9 = 0$ when $x = \pm 3 i$ , so $x^{2} + 9 = (x - 3 i) (x + 3 i)$ . The zeros are $\pm 2$ and the conjugate pair $\pm 3 i$ .
4. D
  That factors $u^{2} + 5 u - 36$ with $u = x^{2}$ and stops there, so its factors are in $x^{2}$ , not $x$ . It is only degree 2, while $p$ has degree 4 and so four linear factors.
No calculator. A polynomial function $p$ has odd degree, and $p (x) > 0$ for every $x > 100$ . Which of the following must be true?
Answer: D.
1. A
  Both ends of a polynomial go the same way only when its degree is even. With odd degree the left end goes the opposite way from the right end.
2. B
  An odd-degree polynomial must cross the $x$ -axis at least once, but it may cross more. For example, $p (x) = x (x - 1) (x - 2)$ is positive for every $x > 100$ and has three real zeros.
3. C
  The outputs do become negative far enough to the left, but not necessarily by $x = - 100$ . For example, $p (x) = x + 200$ is positive for every $x > 100$ , yet $p (- 150) = 50$ .
4. D
  Correct. A nonconstant polynomial's outputs increase or decrease without bound as $x$ increases. They stay positive for $x > 100$ , so $\lim_{x \to \infty} p (x) = \infty$ and the leading coefficient is positive. With odd degree, the left end goes the opposite way, so $\lim_{x \to - \infty} p (x) = - \infty$ .
No calculator. Let $r (x) = \frac{6 x^{4} - x + 2}{5 x + 3 x^{2}}$ . Which function has the same end behavior as $r$ , both as $x$ increases without bound and as $x$ decreases without bound?
Answer: D.
1. A
  Dividing the leading coefficients, $\frac{6}{3}$ , gives a horizontal asymptote only when the degrees match. Here the numerator has degree 4 and the denominator degree 2, so the numerator dominates and $r$ grows without bound.
2. B
  That divides the denominator's leading term by the numerator's, $\frac{3 x^{2}}{6 x^{4}}$ , and flips it the wrong way. The numerator's degree is higher, so $r$ grows without bound instead of approaching 0.
3. C
  That divides by $5 x$ , the first term written in the denominator. The leading term is the one with the highest power, $3 x^{2}$ , even when it is written last.
4. D
  Correct. For inputs of large magnitude each polynomial is dominated by its leading term, so $r (x)$ behaves like $\frac{6 x^{4}}{3 x^{2}} = 2 x^{2}$ . Both $r$ and $2 x^{2}$ increase without bound as $x$ increases and as $x$ decreases without bound.
No calculator. What are all values of $x$ that satisfy $\frac{x}{x - 3} \geq 2$ ?
Answer: B.
1. A
  That multiplies both sides by $x - 3$ without asking its sign: $x \geq 2 x - 6$ . For $x < 3$ the factor is negative and the inequality should flip. Testing $x = 0$ gives $0 \geq 2$ , which is false.
2. B
  Correct. Subtract 2 and combine: $\frac{x - 2 (x - 3)}{x - 3} = \frac{6 - x}{x - 3} \geq 0$ . The numerator is 0 at $x = 6$ and the denominator is 0 at $x = 3$ . The quotient is positive between them and negative outside. It equals 0 at $x = 6$ , which is included, and $x = 3$ is not in the domain.
3. C
  The endpoints are right, but $x = 3$ makes the denominator 0, so $\frac{x}{x - 3}$ has no value there. A zero of the denominator is never part of the solution.
4. D
  That takes the intervals where $\frac{6 - x}{x - 3}$ is negative. Testing $x = 4$ in the original gives $\frac{4}{1} = 4 \geq 2$ , so the values between 3 and 6 are the ones that work.
No calculator. Let $r (x) = \frac{(x - 2)^{m} (x + 1)}{(x - 2)^{3} (x - 5)}$ , where $m$ is a positive integer. For which values of $m$ does the graph of $r$ have a vertical asymptote at $x = 2$ ?
Answer: A.
1. A
  Correct. The factor $x - 2$ appears 3 times in the denominator and $m$ times in the numerator. When $m = 1$ or $m = 2$ the denominator's multiplicity is greater, so for $x \neq 2$ , $r (x) = \frac{x + 1}{(x - 2)^{3 - m} (x - 5)}$ with at least one factor of $x - 2$ left below. Near $x = 2$ the numerator is close to 3 and the denominator is close to 0, so the outputs increase or decrease without bound.
2. B
  With $m = 3$ the multiplicities are equal and every factor of $x - 2$ cancels: for $x \neq 2$ , $r (x) = \frac{x + 1}{x - 5}$ , and near $x = 2$ the outputs approach $\frac{3}{- 3} = - 1$ . They do not grow without bound, so $m = 3$ gives no vertical asymptote.
3. C
  That reverses the comparison. For $m \geq 3$ the numerator has at least as many factors of $x - 2$ as the denominator, so all three below cancel. With $m = 4$ , for example, $r (x) = \frac{(x - 2) (x + 1)}{x - 5}$ for $x \neq 2$ , and the outputs near $x = 2$ approach 0.
4. D
  The denominator is 0 at $x = 2$ for every $m$ , but that gives a vertical asymptote only when the numerator does not cancel all three factors of $x - 2$ . For $m \geq 3$ it cancels all of them.
No calculator. The average rate of change of $f (x) = x^{3}$ from $x$ to $x + 2$ is $\frac{(x + 2)^{3} - x^{3}}{2}$ . Which polynomial is equal to this expression for every real number $x$ ?
Answer: B.
1. A
  That is $(x + 2)^{3} - x^{3}$ , the change in output. The average rate of change still divides it by the change in input, 2.
2. B
  Correct. Row 3 of Pascal's triangle is 1, 3, 3, 1, so $(x + 2)^{3} = x^{3} + 3 \cdot 2 x^{2} + 3 \cdot 2^{2} x + 2^{3} = x^{3} + 6 x^{2} + 12 x + 8$ . Subtracting $x^{3}$ leaves $6 x^{2} + 12 x + 8$ , and dividing by 2 gives $3 x^{2} + 6 x + 4$ .
3. C
  The middle term of the expansion is $3 \cdot 2^{2} x = 12 x$ . Using 2 instead of $2^{2}$ there gives $6 x$ , and halving it gives $3 x$ .
4. D
  That treats $(x + 2)^{3}$ as $x^{3} + 8$ , which leaves out the middle terms of the expansion. Check at $x = 1$ : $\frac{27 - 1}{2} = 13$ , not 4.
No calculator. The graph of $f$ is made of straight segments joining $(0, 0)$ , $(2, 4)$ , $(4, 4)$ and $(8, 0)$ , in that order. The graph of $g$ is made of straight segments joining $(- 2, 1)$ , $(- 1, 7)$ , $(0, 7)$ and $(2, 1)$ , in that order. Which of the following is $g (x)$ ?
Answer: B.
1. A
  That shifts left 2 first and then compresses, which gives $f (2 x + 2) = f (2 (x + 1))$ , a net shift of only 1 to the left. At $x = - 2$ it asks for $f (- 2)$ , which is not defined.
2. B
  Correct. The inputs of $g$ run over 4 units where those of $f$ run over 8, so the horizontal dilation factor is $\frac{1}{2}$ , and $x = - 2$ must map to $f$ 's input 0: $2 (x + 2) = 2 x + 4$ . The outputs run from 1 to 7 where those of $f$ run from 0 to 4, so multiply by $\frac{6}{4} = 1.5$ and add 1. Check: $g (- 1) = 1.5 f (2) + 1 = 7$ and $g (2) = 1.5 f (8) + 1 = 1$ .
3. C
  A horizontal dilation by a factor of $\frac{1}{2}$ multiplies the input by 2, not by 0.5. This version stretches the graph instead, so $g (- 1)$ would be $1.5 f (0.5) + 1 = 2.5$ , not 7.
4. D
  That matches the top, $2 \cdot 4 - 1 = 7$ , but not the bottom: where $f$ is 0 it gives $- 1$ , while the graph of $g$ starts and ends at 1.
No calculator. The arithmetic sequence $a_{n}$ has $a_{0} = 1$ and common difference 3. A second sequence is given by $b_{n} = 2^{a_{n}}$ for every whole number $n$ . Which describes the sequence $b_{n}$ ?
Answer: C.
1. A
  Doubling the common difference does not describe how these terms change. The terms are $2^{1} = 2$ , $2^{4} = 16$ , $2^{7} = 128$ and $2^{10} = 1024$ , and their differences, 14, 112 and 896, are not equal.
2. B
  That multiplies the base 2 by the common difference 3. Adding 3 to an exponent multiplies the power by $2^{3}$ , not by $2 \cdot 3$ : $\frac{16}{2} = 8$ .
3. C
  Correct. The exponents $a_{n} = 1 + 3 n$ are 1, 4, 7, 10, so the terms are 2, 16, 128 and 1024. Each step adds 3 to the exponent, and $2^{k + 3} = 8 \cdot 2^{k}$ , so each term is 8 times the one before: $b_{n} = 2 \cdot 8^{n}$ .
4. D
  The changing differences rule out arithmetic, but not geometric. Divide successive terms: $\frac{16}{2} = \frac{128}{16} = \frac{1024}{128} = 8$ , a common ratio.
No calculator. The table gives values of a function $f$ at five inputs.
$\begin{matrix} x & 0 & 1 & 3 & 4 & 6 \\ f (x) & 5 & 10 & 40 & 80 & 320 \end{matrix}$
Could $f$ be an exponential function of the form $f (x) = a b^{x}$ ?
Answer: C.
1. A
  The inputs are not equally spaced. The ratio 4 comes from the steps of length 2, from 1 to 3 and from 4 to 6, and two doublings make a factor of 4.
2. B
  Equal differences are the test for a linear function. An exponential function has unequal differences, so this cannot rule it out.
3. C
  Correct. Over each interval of length 1 in the table, 0 to 1 and 3 to 4, the outputs double. Over each interval of length 2, 1 to 3 and 4 to 6, they are multiplied by 4. That fits $f (x) = 5 \cdot 2^{x}$ , which gives every value in the table.
4. D
  Many functions are positive and increasing without being exponential, such as $x^{2} + 5$ for $x \geq 0$ . What marks an exponential function is equal ratios over equal-length intervals.
No calculator. A shop prints T-shirts. $N (h)$ is the number of shirts printed in the first $h$ hours of a shift. $I (n)$ is the number of liters of ink used to print $n$ shirts. $C (v)$ is the cost, in dollars, of $v$ liters of ink. Which expression gives the cost of the ink used in the first $h$ hours of a shift?
Answer: A.
1. A
  Correct. Hours go into $N$ and give shirts. Shirts are what $I$ takes, and it gives liters. Liters are what $C$ takes, and it gives dollars. The function that acts first is written closest to $h$ .
2. B
  The functions are the right ones, but in reverse order. In $N (I (C (h)))$ , $C$ would act first and treat a number of hours as liters of ink.
3. C
  After $N$ gives a number of shirts, $C$ would treat that number as liters of ink. Each function must receive the kind of quantity it takes.
4. D
  $I$ takes a number of shirts, not a number of hours, so it cannot act on $h$ first.
No calculator. The function $f$ is defined only for $0 \leq x \leq 6$ , and its graph is made of straight segments joining $(0, 1)$ , $(3, 5)$ and $(6, 2)$ . Let $g (x) = 6 - x$ . Which describes the graph of $h (x) = f (g (x))$ for $0 \leq x \leq 6$ ?
Answer: D.
1. A
  That is the graph of $f$ itself. Composing with $g$ changes which input of $f$ is used: $h (0) = f (6) = 2$ , not $f (0)$ .
2. B
  Those are values of $g (f (x)) = 6 - f (x)$ , with the functions applied in the other order. In $f (g (x))$ , $g$ acts first and $f$ acts on its output.
3. C
  That reflects the graph of $f$ over the $x$ -axis, which is what $- f (x)$ does. Here $g$ changes the input of $f$ , not its output, so the outputs of $h$ are outputs of $f$ and stay between 1 and 5.
4. D
  Correct. $h (0) = f (6) = 2$ , $h (3) = f (3) = 5$ and $h (6) = f (0) = 1$ . Between these, $g$ runs straight back across each segment of $f$ , so the graph of $h$ is the graph of $f$ reflected over the line $x = 3$ .
No calculator. The function $f$ is defined only for $x = 1, 2, 3, 4, 5, 6$ , and the table gives all of its values.
$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 & 6 \\ f (x) & 4 & 6 & 3 & 1 & 2 & 5 \end{matrix}$
What are all values of $x$ for which $f (x) = f^{- 1} (x)$ ?
Answer: B.
1. A
  3 is the only input with $f (x) = x$ , where the graph meets the line $y = x$ . But $f$ and $f^{- 1}$ can also agree at a pair of points that are reflections of each other: $f (1) = 4$ and $f (4) = 1$ .
2. B
  Correct. Reverse the pairs to get $f^{- 1}$ : $f^{- 1} (1) = 4$ , $f^{- 1} (2) = 5$ , $f^{- 1} (3) = 3$ , $f^{- 1} (4) = 1$ , $f^{- 1} (5) = 6$ , $f^{- 1} (6) = 2$ . Compare with $f$ : at 1 both are 4, at 3 both are 3, at 4 both are 1. At 2, 5 and 6 they differ.
3. C
  That treats $f^{- 1} (x)$ as $\frac{1}{f (x)}$ , which equals $f (x)$ only where $f (x) = 1$ . The inverse reverses each input-output pair; it is not a reciprocal.
4. D
  Those two do work, since $f$ swaps 1 and 4. But $f (3) = 3$ means $f^{- 1} (3) = 3$ as well, so $x = 3$ belongs too.
No calculator. If $\log_{b} 81 = \frac{4}{3}$ , what is the value of $b$ ?
Answer: D.
1. A
  That multiplies 81 by $\frac{4}{3}$ . A logarithm is an exponent: $\log_{b} 81 = \frac{4}{3}$ means $b^{4 / 3} = 81$ .
2. B
  $3^{4} = 81$ , so $\log_{3} 81 = 4$ , not $\frac{4}{3}$ . The 3 in the denominator of the exponent still has to be accounted for.
3. C
  That divides 81 by $\frac{4}{3}$ , treating the logarithm as a product. In exponential form the equation is $b^{4 / 3} = 81$ , and ${60.75}^{4 / 3}$ is far more than 81.
4. D
  Correct. In exponential form, $b^{4 / 3} = 81$ , so $b = 81^{3 / 4}$ . The fourth root of 81 is 3, and $3^{3} = 27$ . Check: $27^{4 / 3} = 3^{4} = 81$ .
No calculator. Let $f (x) = 5 (0.4)^{x}$ . Which describes the graph of the logarithmic function $f^{- 1}$ , the inverse of $f$ , on its domain?
Answer: C.
1. A
  That is the shape of $\log_{2} x$ , the inverse of a growing exponential. This $f$ is decreasing, and an inverse function goes the same direction as the function it undoes.
2. B
  The direction is right, but reflecting a decreasing graph over $y = x$ does not flip its concavity. Points on $f^{- 1}$ such as $(0.8, 2)$ , $(2, 1)$ and $(5, 0)$ have slopes about $- 0.83$ and then $- 0.33$ , which increase.
3. C
  Correct. $f$ is decreasing and concave up, through $(0, 5)$ , $(1, 2)$ and $(2, 0.8)$ . Its inverse has the points $(0.8, 2)$ , $(2, 1)$ and $(5, 0)$ . The inverse decreases, and its slopes go from $\frac{- 1}{1.2} \approx - 0.83$ to $\frac{- 1}{3} \approx - 0.33$ , rising toward 0, so it is concave up.
4. D
  A function and its inverse change in the same direction. Since $f$ falls as $x$ increases, so does $f^{- 1}$ .
No calculator. What are all values of $x$ that satisfy $\log_{1 / 2} (x - 1) > - 2$ ?
Answer: A.
1. A
  Correct. The logarithm needs $x - 1 > 0$ , so $x > 1$ . Because the base $\frac{1}{2}$ is less than 1, $\log_{1 / 2}$ is decreasing, so $\log_{1 / 2} (x - 1) > - 2$ means $x - 1 < {(\frac{1}{2})}^{- 2} = 4$ , or $x < 5$ .
2. B
  That keeps the inequality's direction when undoing the logarithm. With a base less than 1, the logarithm is decreasing, so larger inputs give smaller outputs and the direction reverses.
3. C
  The upper bound is right, but a logarithm takes only positive inputs. For $x \leq 1$ , $x - 1$ is not positive and $\log_{1 / 2} (x - 1)$ is not defined.
4. D
  That uses ${(\frac{1}{2})}^{- 2} = \frac{1}{4}$ . A negative exponent means the reciprocal: ${(\frac{1}{2})}^{- 2} = 2^{2} = 4$ .
No calculator. The populations of towns A and B are recorded each year and graphed on the same semi-log plot, with the vertical axis scaled by base 10 logarithms. The points for each town lie on a straight line. The two lines are parallel, and town A's line is above town B's. Which conclusion is supported by the plot?
Answer: B.
1. A
  On a semi-log plot the growth factor shows in the slope, not the height. Parallel lines have the same slope, so the towns grow by the same percent each year. A higher line only means a larger population.
2. B
  Correct. The vertical distance between the lines is $\log A - \log B = \log \frac{A}{B}$ , and parallel lines keep that distance constant. So $\frac{A}{B}$ never changes: both towns grow by the same factor each year.
3. C
  On a logarithmic axis a constant gap means a constant ratio, not a constant difference. With the same growth factor, the larger town gains more people each year, so the difference grows.
4. D
  A straight line on a semi-log plot means the logarithm of the population is linear, so the population itself is exponential.
No calculator. An angle $θ$ in standard position has $\cos θ = - \frac{20}{29}$ and $\tan θ > 0$ . Which point lies on the terminal ray of $θ$ ?
Answer: B.
1. A
  That point is in Quadrant II, where $\tan θ = \frac{21}{- 20}$ is negative. The tangent is positive here.
2. B
  Correct. Cosine is negative and tangent is positive only in Quadrant III, so both coordinates are negative. The point is $\sqrt{20^{2} + 21^{2}} = 29$ from the origin, so $\cos θ = - \frac{20}{29}$ , and $\tan θ = \frac{- 21}{- 20} > 0$ .
3. C
  That point is in Quadrant III, but its cosine is $- \frac{21}{29}$ . The cosine is the $x$ -coordinate over the distance from the origin, so the $x$ -coordinate must be $- 20$ .
4. D
  That point is in Quadrant IV, where the cosine is positive. Here $\cos θ$ is negative.
No calculator. As $θ$ increases from $\frac{π}{3}$ to $\frac{4 π}{3}$ , which gives every value taken by $\cos θ$ ?
Answer: D.
1. A
  Those are the values at the two ends of the interval, $\cos \frac{π}{3} = \frac{1}{2}$ and $\cos \frac{4 π}{3} = - \frac{1}{2}$ . In between, at $θ = π$ , the point on the unit circle reaches $(- 1, 0)$ , so cosine gets down to $- 1$ .
2. B
  Cosine reaches 1 only at multiples of $2 π$ , and none lies in this interval. The greatest value here is $\frac{1}{2}$ , at the start.
3. C
  Those are the values of $\sin θ$ on this interval, which rises to 1 at $\frac{π}{2}$ and ends at $- \frac{\sqrt{3}}{2}$ . Cosine is the $x$ -coordinate of the point on the unit circle, not the $y$ -coordinate.
4. D
  Correct. Cosine is the $x$ -coordinate of the point on the unit circle. It starts at $\frac{1}{2}$ , decreases to $- 1$ at $θ = π$ , then increases to $- \frac{1}{2}$ at $\frac{4 π}{3}$ . Every value from $- 1$ to $\frac{1}{2}$ is taken.
No calculator. Which of the following is an odd function?
Answer: C.
1. A
  Cosine is even: its graph is symmetric over the $y$ -axis, so $3 \cos (- 2 θ) = 3 \cos (2 θ)$ . An odd function needs $f (- θ) = - f (θ)$ .
2. B
  It is written with sine, but $\sin (θ + \frac{π}{2}) = \cos θ$ , which is even. The phase shift moved the graph off its symmetry about the origin.
3. C
  Correct. $f (- θ) = 2 \sin (- 3 θ) = - 2 \sin (3 θ) = - f (θ)$ , because sine is odd. Dilations keep the graph symmetric about the origin.
4. D
  Adding 2 moves the graph up, so it is no longer symmetric about the origin: $f (\frac{π}{2}) = 3$ but $f (- \frac{π}{2}) = 1$ , not $- 3$ .
No calculator. The table gives values of a sinusoidal function $g$ .
$\begin{matrix} θ & 0 & \frac{π}{4} & \frac{π}{2} & \frac{3 π}{4} & π \\ g (θ) & 3 & 1 & - 1 & 1 & 3 \end{matrix}$
Which of the following could define $g$ ?
Answer: A.
1. A
  Correct. The table runs from 3 down to $- 1$ and back, so the midline is $y = 1$ and the amplitude is 2. A maximum at 0 and the next at $π$ fit a cosine with period $π$ , so $b = 2$ . Check: $2 \cos \frac{π}{2} + 1 = 1$ and $2 \cos π + 1 = - 1$ .
2. B
  That has period $\frac{2 π}{1 / 2} = 4 π$ , not $π$ . Its value at $θ = π$ is $2 \cos \frac{π}{2} + 1 = 1$ , not 3.
3. C
  That takes the maximum, 3, as the amplitude and leaves out the midline. It gives $- 3$ at $\frac{π}{2}$ , but the table's least value is $- 1$ .
4. D
  A sine graph with no phase shift starts on its midline, so this gives 1 at $θ = 0$ . The table starts at a maximum, which is where cosine starts.
No calculator. Angles are in radians. Which of the following has the greatest value?
Answer: A.
1. A
  Correct. Angles of 2 and 3 radians are in Quadrant II, where tangent is negative. Tangent has period $π$ , so $\tan 4 = \tan (4 - π)$ , and $4 - π \approx 0.86$ . Tangent increases on $- \frac{π}{2} < θ < \frac{π}{2}$ , and $0.86 < 1$ , so $\tan 4 < \tan 1$ .
2. B
  2 is close to $\frac{π}{2} \approx 1.57$ , but on the far side of the asymptote. Just past $\frac{π}{2}$ the terminal ray is in Quadrant II and its slope is negative.
3. C
  3 is just under $π$ , so its terminal ray is in Quadrant II and its slope is a small negative number.
4. D
  Tangent increases only between consecutive asymptotes, not across them. Since $\tan 4 = \tan (4 - π)$ and $4 - π \approx 0.86$ is less than 1, $\tan 4$ is less than $\tan 1$ .
No calculator. A kite flies on a straight, taut string 50 meters long. The string is held at ground level, and the kite is $h$ meters above the ground, where $0 \leq h \leq 50$ . Which expression gives the angle $θ$ , in radians, between the string and the ground?
Answer: B.
1. A
  $\frac{h}{50}$ is the side opposite the angle at the ground over the hypotenuse, which is a sine. The arccosine of that ratio is the angle between the string and a vertical line.
2. B
  Correct. The string, the ground and the vertical drop from the kite make a right triangle with hypotenuse 50 and the side opposite $θ$ equal to $h$ . So $\sin θ = \frac{h}{50}$ , and since $0 \leq θ \leq \frac{π}{2}$ , arcsine gives $θ$ directly.
3. C
  Tangent is opposite over adjacent. The 50 meters is the string, the hypotenuse, not the distance along the ground.
4. D
  The ratio is upside down. $\frac{50}{h}$ is at least 1 for every height below 50 meters, outside the inputs arcsine accepts.
No calculator. What are all values of $x$ with $0 \leq x < 2 π$ that satisfy $\sin x > \cos x$ ?
Answer: C.
1. A
  On that interval $\sin x > \cos x$ does hold, but it keeps holding after $\frac{3 π}{4}$ : at $x = π$ , $\sin x = 0 > - 1 = \cos x$ . The two are equal again only at $\frac{5 π}{4}$ .
2. B
  Those are the values where $\cos x > \sin x$ . Check $x = 0$ : $\sin 0 = 0$ is less than $\cos 0 = 1$ .
3. C
  Correct. $\sin x = \cos x$ where the point on the unit circle has equal coordinates, at $\frac{π}{4}$ and $\frac{5 π}{4}$ . Between them the point is above the line $y = x$ , so its $y$ -coordinate is larger. Check $x = \frac{π}{2}$ : $1 > 0$ .
4. D
  That is where $\sin x > 0 > \cos x$ , but $\sin x$ can beat $\cos x$ while both are positive, as at $x = \frac{π}{3}$ , or both negative, as at $x = \frac{7 π}{6}$ .
No calculator. As $θ$ increases from just above $\frac{π}{2}$ to $π$ , how does $\sec θ$ change?
Answer: C.
1. A
  That describes $\cos θ$ on this interval. Its reciprocal behaves differently: when $\cos θ$ is close to 0, $\sec θ$ is far from 0.
2. B
  On this interval $\cos θ$ is negative, so its reciprocal is negative too.
3. C
  Correct. Here $\cos θ$ falls from just below 0 to $- 1$ . Just past $\frac{π}{2}$ , $\cos θ$ is a negative number near 0, so $\sec θ = \frac{1}{\cos θ}$ is a negative number of very large magnitude, such as $\frac{1}{- 0.01} = - 100$ . As $\cos θ$ moves toward $- 1$ , $\sec θ$ rises to $\frac{1}{- 1} = - 1$ .
4. D
  The values are right but the direction is reversed. $\sec θ$ is $- 1$ at $θ = π$ , the end of the interval, and very negative near $\frac{π}{2}$ , the start, so it increases.
No calculator. Which expression is equal to $\frac{\sec^{2} θ - 1}{\sec^{2} θ}$ for every $θ$ where the expression is defined?
Answer: D.
1. A
  Splitting the fraction gives $1 - \frac{1}{\sec^{2} θ} = 1 - \cos^{2} θ$ . Stopping at $\cos^{2} θ$ drops the subtraction from 1.
2. B
  $\sec^{2} θ - 1 = \tan^{2} θ$ is only the numerator. The whole numerator is still divided by $\sec^{2} θ$ .
3. C
  That cancels $\sec^{2} θ$ from one term of the numerator. A factor can cancel only from the whole numerator, not from part of a difference.
4. D
  Correct. By the Pythagorean identity, $\sec^{2} θ - 1 = \tan^{2} θ$ , so the expression is $\frac{\tan^{2} θ}{\sec^{2} θ} = \frac{\sin^{2} θ}{\cos^{2} θ} \cdot \cos^{2} θ = \sin^{2} θ$ . Or split it: $1 - \cos^{2} θ = \sin^{2} θ$ .
No calculator. A point $P$ has polar coordinates $(r, θ)$ , with $r \neq 0$ . Which polar coordinates name the reflection of $P$ over the $y$ -axis, whatever the values of $r$ and $θ$ ?
Answer: A.
1. A
  Correct. $P$ has rectangular coordinates $(r \cos θ, r \sin θ)$ , so its reflection over the $y$ -axis is $(- r \cos θ, r \sin θ)$ . The pair $(- r, - θ)$ gives $x = - r \cos (- θ) = - r \cos θ$ and $y = - r \sin (- θ) = r \sin θ$ , which is that point. It is another name for $(r, π - θ)$ .
2. B
  This gives $x = r \cos (- θ) = r \cos θ$ and $y = r \sin (- θ) = - r \sin θ$ : the same $x$ -coordinate and the opposite $y$ -coordinate. That is the reflection over the $x$ -axis.
3. C
  A negative $r$ with the same angle sends the point to the opposite ray, $(- r \cos θ, - r \sin θ)$ . Both coordinates change sign, which is the reflection through the origin.
4. D
  The angle $π - θ$ alone would reflect $P$ over the $y$ -axis, but the negative $r$ then sends the point to the opposite ray: $x = - r \cos (π - θ) = r \cos θ$ and $y = - r \sin (π - θ) = - r \sin θ$ . That is the reflection over the $x$ -axis.

Section I Part B: Multiple Choice (graphing calculator required)

13 questions · 40 minutes · Graphing calculator required

Graphing calculator required. A hiker climbs a steady trail to the top of a ridge and starts straight down the other side. The table gives her elevation, $E (t)$ meters above sea level, $t$ hours after she sets out.
$\begin{matrix} t & 5.8 & 5.9 & 5.95 & 6 & 6.05 & 6.1 & 6.2 \\ E (t) & 1259.4 & 1267.8 & 1271.9 & 1276.0 & 1270.5 & 1265.0 & 1254.1 \end{matrix}$
A student estimates the rate of change of $E$ at $t = 6$ with the average rate of change from $t = 5.9$ to $t = 6.1$ . Which statement about that estimate is supported by the table?
Answer: B.
1. A
  A centered interval gives a good estimate only when the average rates over small intervals on both sides settle toward one value. Here the elevation rises at about 82 meters per hour up to $t = 6$ and falls at about 110 meters per hour after it, so the average from 5.9 to 6.1, $\frac{1265.0 - 1267.8}{0.2} = - 14$ , blends two different rates and matches neither.
2. B
  Correct. Intervals ending at $t = 6$ give $\frac{1276.0 - 1259.4}{0.2} = 83$ , $\frac{1276.0 - 1267.8}{0.1} = 82$ and $\frac{1276.0 - 1271.9}{0.05} = 82$ . Intervals starting at $t = 6$ give $\frac{1254.1 - 1276.0}{0.2} = - 109.5$ , $\frac{1265.0 - 1276.0}{0.1} = - 110$ and $\frac{1270.5 - 1276.0}{0.05} = - 110$ . The two sides settle near different values, which is what happens when she tops the ridge and starts down, so small intervals do not single out one rate at $t = 6$ . The value $- 14$ is a blend of the climbing rate and the descending rate, not the rate at any moment.
3. C
  The table answers this directly. From $t = 5.95$ to $t = 6.05$ the average rate of change is $\frac{1270.5 - 1271.9}{0.1} = - 14$ , the same as before. Shrinking a centered interval does not help here, because each side keeps its own rate, about 82 on the left and about $- 110$ on the right.
4. D
  From $t = 6$ to $t = 6.1$ the average rate is $- 110$ meters per hour, the rate just after she tops the ridge, but that ignores the climb at about 82 meters per hour just before. A rate at a point is estimated from small intervals containing the point, and here the intervals on the two sides disagree.
Graphing calculator required. A crew pumps floodwater out of a basement. An engineer models the depth of the water, in centimeters, $t$ minutes after the pumping starts by $D (t) = 0.0135 t^{2} - 1.62 t + 48.3$ . Which restriction on the domain of $D$ fits this situation, with a correct reason?
Answer: A.
1. A
  Correct. Graphing $y = D (t)$ shows it falling to 0 at $t \approx 55.286$ , staying below 0 until $t \approx 64.714$ , and rising after that. A depth cannot be negative: the basement is empty at $t \approx 55.286$ , so the model describes the water only from $t = 0$ until then.
2. B
  The least value of the model is $D (60) = - 0.3$ , a negative depth. The model is already below 0 from $t \approx 55.286$ to $t = 60$ , so the domain has to stop sooner.
3. C
  Between its two zeros the model is negative, for example $D (60) = - 0.3$ , so this domain includes about 9.4 minutes of negative depths. The water is gone at the first zero, $t \approx 55.286$ .
4. D
  Time does start at 0, but this keeps every later time too. For large $t$ the term $0.0135 t^{2}$ makes the model rise again, as if the basement refilled while the pumps run, and it gives negative depths just before that.
Questions 3 to 4 refer to the following.
A company designs a closed cylindrical can that holds 450 cubic centimeters. If the can's radius is $r$ centimeters, its height must be $\frac{450}{π r^{2}}$ centimeters, and the area of metal it uses for the top, the bottom and the side is $S (r) = 2 π r^{2} + \frac{900}{r}$ square centimeters, for $r > 0$ . The model ignores any metal lost at the seams.

Graphing calculator required. The company wants each can to use at most 400 square centimeters of metal. For which radii $r$ does the can use 400 square centimeters of metal or less, that is, $S (r) \leq 400$ ?
Answer: D.
1. A
  These are the radii where the can uses at least 400 square centimeters, $S (r) \geq 400$ . Near $r = 0$ the term $\frac{900}{r}$ is huge, and for large $r$ the term $2 π r^{2}$ is huge, so both ends of the domain use too much metal. The radii that work lie between the two crossings.
2. B
  This finds only the first crossing. In a window that stops near $r = 5$ , the graph of $S$ drops below 400 at $r \approx 2.494$ and seems to stay there, but $2 π r^{2}$ grows without bound, and $S$ rises back through 400 at $r \approx 6.434$ .
3. C
  This solves the inequality as if every real $r$ were allowed. For $- 8.928 \leq r < 0$ the term $\frac{900}{r}$ is negative, so $S (r) \leq 400$ there, but a radius is a length and must be positive. The context restricts the domain to $r > 0$ .
4. D
  Correct. $S (r) - 400 = \frac{2 π r^{3} - 400 r + 900}{r}$ . For $r > 0$ its sign matches the numerator, whose zeros are about $- 8.928$ , $2.494$ and $6.434$ ; only the last two are radii. Graphing $y = S (r)$ and $y = 400$ shows $S (r) \leq 400$ exactly between those two intersections, so $2.494 \leq r \leq 6.434$ .
Uses the material above question 3.

Graphing calculator required. According to the model, what is the least amount of metal, in square centimeters, that a can holding 450 cubic centimeters can use?
Answer: B.
1. A
  4.153 centimeters is the radius at which the least metal is used, the input of the minimum. The question asks for the amount of metal, the output $S (4.153) \approx 325.080$ .
2. B
  Correct. For $r > 0$ the graph of $S (r) = 2 π r^{2} + \frac{900}{r}$ falls and then rises, with its minimum at $r \approx 4.153$ . The minimum value is $S \approx 325.080$ square centimeters, so no can that holds 450 cubic centimeters uses less metal than that.
3. C
  325.531 is $S (4) = 32 π + 225$ , the smallest value in a table of whole-number radii. The minimum lies between whole numbers, at $r \approx 4.153$ , where $S \approx 325.080$ is a little lower.
4. D
  8.306 centimeters is the height of the best can, $\frac{450}{π (4.153)^{2}}$ , which equals its diameter. It is a length, not an amount of metal; the least area is $S (4.153) \approx 325.080$ square centimeters.
Graphing calculator required. A city counts its public charging ports for electric cars on January 1 of each year. With $t$ the number of years after January 1, 2019, the counts were 46, 61, 77, 102, 129, 168 and 214 for $t = 0, 1, 2, \dots, 6$ . A planner fits an exponential regression $N (t) = a b^{t}$ to these counts and assumes the same growth pattern continues. Using the regression as your calculator stores it, how many years does the model take to double the number of ports?
Answer: D.
1. A
  This is $\frac{\ln b}{\ln 2}$ , the ratio upside down. Doubling means $b^{T} = 2$ , so $T = \frac{\ln 2}{\ln b}$ . A doubling time of about a third of a year does not fit the data, which go from 46 to 102 in three years.
2. B
  This is $\frac{\ln 2}{b - 1}$ , which treats the yearly growth rate, about 0.291, as if it were the constant $k$ in $e^{k t}$ . The yearly growth factor is $b \approx 1.291$ , so the doubling time solves $b^{T} = 2$ .
3. C
  This solves $a b^{T} = 92$ , the time for the model to reach twice the first count, measured from $t = 0$ . The model starts at $a \approx 46.608$ , not 46, so 92 is less than double its starting value. The doubling time comes from $b^{T} = 2$ and is the same from any starting year.
4. D
  Correct. The regression gives $a \approx 46.608$ and $b \approx 1.291$ . The model doubles when $b^{T} = 2$ , so $T = \frac{\ln 2}{\ln b} \approx 2.713$ years, whatever year you start from. The number of ports doubles about every 2.7 years.
Graphing calculator required. Let $f (x) = 2^{x} + 3^{x}$ for all real numbers $x$ . Because $f$ is increasing, it has an inverse function $f^{- 1}$ . What is the value of $f^{- 1} (20)$ ?
Answer: C.
1. A
  This is $\log_{6} 20$ , from treating $2^{x} + 3^{x}$ as $6^{x}$ , as if the bases multiply. They do not: $2^{2} + 3^{2} = 13$ , while $6^{2} = 36$ .
2. B
  This is $\log_{5} 20$ , from adding the bases. $2^{x} + 3^{x}$ is not $5^{x}$ : $2^{2} + 3^{2} = 13$ , while $5^{2} = 25$ .
3. C
  Correct. $f^{- 1} (20)$ is the input that $f$ sends to 20, so solve $2^{x} + 3^{x} = 20$ . No algebraic step isolates $x$ , but graphing $y = 2^{x} + 3^{x}$ and $y = 20$ gives one intersection, at $x \approx 2.439$ . Check: $2^{2.439} + 3^{2.439} \approx 20$ .
4. D
  This is $\log_{3} 20$ , which drops the $2^{x}$ term and solves $3^{x} = 20$ . Then $f (2.727) = 2^{2.727} + 20 \approx 26.620$ , too big. With both terms counted, the input is smaller.
Graphing calculator required. A classroom's air holds 1800 parts per million (ppm) of carbon dioxide when a ventilation fan is switched on at $t = 0$ . With the fan running and nobody in the room, the level $t$ minutes later is modeled by $C (t) = 420 + 1380 (0.91)^{t}$ ppm. The model assumes the level falls toward the outdoor level of 420 ppm. According to the model, during which time interval is the level at most 1000 ppm and at least 600 ppm?
Answer: A.
1. A
  Correct. $C (t) \leq 1000$ when $1380 (0.91)^{t} \leq 580$ , which first happens at $t = \frac{\ln (580 / 1380)}{\ln 0.91} \approx 9.191$ . $C (t) \geq 600$ while $1380 (0.91)^{t} \geq 180$ , which last holds at $t = \frac{\ln (180 / 1380)}{\ln 0.91} \approx 21.598$ . Since $C$ is decreasing, the level is in that range from about 9.191 to 21.598 minutes.
2. B
  This drops the 420 and solves $1380 (0.91)^{t} = 1000$ and $1380 (0.91)^{t} = 600$ . The level is 420 plus the decaying part, so subtract 420 from 1000 and from 600 before dividing by 1380.
3. C
  This solves $1800 (0.91)^{t} = 1000$ and $1800 (0.91)^{t} = 600$ , treating the starting level as the multiplier of ${0.91}^{t}$ . In this model only the 1380 ppm above the outdoor level decays; the 420 stays.
4. D
  This takes the common logarithm of the ratio but the natural logarithm of 0.91, as in $\frac{\log (580 / 1380)}{\ln 0.91}$ . Both logarithms must use the same base; with matching bases the answer is the same whichever base you choose.
Graphing calculator required. The balance of a savings account $t$ years after it is opened is $B (t) = 2500 (1.043)^{t}$ dollars. The same function can be written as $B (t) = 2500 e^{k t}$ . What is the value of $k$ , to four decimal places?
Answer: B.
1. A
  0.043 is the yearly growth rate, 4.3%. Writing $e^{0.043 t}$ would give a yearly factor of $e^{0.043} \approx 1.0439$ , slightly more than 1.043. The exponent $k$ has to satisfy $e^{k} = 1.043$ .
2. B
  Correct. $1.043 = e^{\ln 1.043}$ , so $(1.043)^{t} = e^{(\ln 1.043) t}$ and $k = \ln 1.043 \approx 0.0421$ . Check at $t = 10$ : with $k$ unrounded, $2500 (1.043)^{10}$ and $2500 e^{10 k}$ are both about 3808.76 dollars.
3. C
  This is $\log 1.043$ , the base 10 logarithm from the calculator's log key. Undoing a power of $e$ takes the natural logarithm, $\ln 1.043$ .
4. D
  This is $e^{1.043}$ , which runs the conversion backward. The value of $k$ has to satisfy $e^{k} = 1.043$ , so $k = \ln 1.043$ , and a growth factor just above 1 gives a $k$ just above 0.
Graphing calculator required. A nursery models the height of a seedling, in centimeters, $d$ days after it sprouts by $H (d) = 2.1 + 3.6 \ln d$ , for $d \geq 1$ . A grower wants the same model in terms of $w$ , the number of weeks after the seedling sprouts, so $d = 7 w$ . Which gives the model in terms of $w$ ?
Answer: C.
1. A
  That treats $\ln (7 w)$ as $7 \ln w$ . The power property moves an exponent out front, but here 7 multiplies the input; it is not an exponent. The product property gives $\ln (7 w) = \ln 7 + \ln w$ .
2. B
  That uses $d = \frac{w}{7}$ , dividing by 7 instead of multiplying. A number of days is 7 times the number of weeks, so $\ln (7 w) = \ln 7 + \ln w$ , and $3.6 \ln 7$ is added, not subtracted.
3. C
  Correct. Substitute $d = 7 w$ and use the product property: $H = 2.1 + 3.6 \ln (7 w) = 2.1 + 3.6 \ln 7 + 3.6 \ln w$ . Since $3.6 \ln 7 \approx 7.005$ , $H \approx 9.105 + 3.6 \ln w$ . Check at 2 weeks: $9.105 + 3.6 \ln 2 \approx 11.601$ , and 14 days gives $2.1 + 3.6 \ln 14 \approx 11.601$ .
4. D
  That adds $\ln 7 \approx 1.946$ but forgets that it is multiplied by 3.6 like the rest of $\ln (7 w)$ . The constant is $2.1 + 3.6 \ln 7 \approx 9.105$ .
Graphing calculator required. In a steady wind, the top of a tall tower sways east and west. Its displacement east of its resting position, in centimeters, $t$ seconds after a sensor starts recording, is modeled by $D (t) = 12 \sin (b (t + c))$ . One full sway takes 5.2 seconds, so $b = \frac{2 π}{5.2}$ . The sensor reads $D (1) = 7.957$ and $D (2.5) = 6.817$ . Which value of $c$ , with $- 2.6 \leq c < 2.6$ , gives a model that agrees with both readings?
Answer: A.
1. A
  Correct. The first reading gives $\sin (b (1 + c)) = \frac{7.957}{12}$ , so $b (1 + c) \approx 0.7249$ or $b (1 + c) \approx π - 0.7249 \approx 2.4167$ . Dividing by $b \approx 1.2083$ gives $c \approx - 0.4$ or $c \approx 1.0$ . The second reading decides: with $c = - 0.4$ , $D (2.5) = 12 \sin (2.1 b) \approx 6.817$ , a match, while $c = 1.0$ gives about $- 10.625$ .
2. B
  With $c = 1.0$ the model gives $D (1) \approx 7.957$ , so it fits the first reading, but with the tower on its way back west. At $t = 2.5$ it gives about $- 10.625$ , not 6.817. One reading allows two phase shifts in each period; the second reading picks one of them.
3. C
  This comes from writing $t - c$ in place of $t + c$ , which flips the sign of $c$ . With $c = 0.4$ the model gives $D (1) \approx 11.913$ , not 7.957.
4. D
  This divides by 5.2 instead of by $b$ , treating the period as the value of $b$ . With $b = \frac{2 π}{5.2}$ the value $c = - 0.861$ gives $D (1) \approx 2.006$ , far from 7.957.
Graphing calculator required. A spotlight hangs 8 meters in front of the back wall of a stage and turns left and right at the height of a point $P$ on the wall. When the light is turned $θ$ radians from pointing straight at $P$ , its beam hits the wall $x = 8 \tan θ$ meters to the right of $P$ , for $- \frac{π}{2} < θ < \frac{π}{2}$ . A negative $x$ means left of $P$ . A banner runs along the wall at that height from 1.5 meters left of $P$ to 6 meters right of $P$ . For which values of $θ$ does the beam hit the banner?
Answer: C.
1. A
  These come from $\sin θ = \frac{- 1.5}{8}$ and $\sin θ = \frac{6}{8}$ , which treats the 8 meters as the length of the beam. The 8 meters runs straight to the wall, next to the angle, and $x$ is opposite the angle, so the ratio is a tangent.
2. B
  These come from $\tan θ = \frac{8}{- 1.5}$ and $\tan θ = \frac{8}{6}$ , the ratio upside down. Since $x = 8 \tan θ$ , the tangent is $\frac{x}{8}$ .
3. C
  Correct. The beam is on the banner when $- 1.5 \leq 8 \tan θ \leq 6$ , that is, $- 0.1875 \leq \tan θ \leq 0.75$ . Tangent is increasing on $- \frac{π}{2} < θ < \frac{π}{2}$ , so taking the arctangent keeps the order: $\arctan (- 0.1875) \approx - 0.185 \leq θ \leq \arctan 0.75 \approx 0.644$ .
4. D
  These are the angles that put the beam at or beyond the ends of the banner, $8 \tan θ \leq - 1.5$ or $8 \tan θ \geq 6$ . Because tangent increases on this interval, the angles that hit the banner lie between the two endpoints, not outside them.
Graphing calculator required. The polar function $f$ is given by $f (θ) = 0.9 + 2.4 \cos θ$ for $0 \leq θ < 2 π$ . Which list gives every point on the polar graph of $r = f (θ)$ that is 1.3 units from the origin, in rectangular coordinates $(x, y)$ ?
Answer: B.
1. A
  The points $(- 0.217, \pm 1.282)$ are the first two points reflected through the origin. A point $(r, θ)$ on the graph does not put $(- r, θ)$ on the graph too: $f$ gives one signed radius for each angle, and $f (θ) = - 1.3$ happens at different angles.
2. B
  Correct. The distance from the origin is $| r |$ , so solve $f (θ) = 1.3$ and $f (θ) = - 1.3$ . The first gives $\cos θ = \frac{0.4}{2.4}$ , so $θ \approx 1.403$ or $θ \approx 4.880$ , and the points $(1.3 \cos θ, 1.3 \sin θ)$ are $(0.217, \pm 1.282)$ . The second gives $\cos θ = - \frac{2.2}{2.4}$ , so $θ \approx 2.730$ or $θ \approx 3.553$ . With $r = - 1.3$ the points are $(- 1.3 \cos θ, - 1.3 \sin θ)$ , which are $(1.192, - 0.520)$ and $(1.192, 0.520)$ , on the inner loop.
3. C
  This solves only $f (θ) = 1.3$ . The distance from the origin is $| r |$ , and $f (θ) = - 1.3$ also has solutions, $θ \approx 2.730$ and $θ \approx 3.553$ , which give two more points, on the inner loop.
4. D
  This finds the right angles for $r = - 1.3$ but plots those points on the terminal ray, at $(1.3 \cos θ, 1.3 \sin θ)$ . A negative $r$ puts the point on the opposite ray, which flips the signs of both coordinates, giving $(1.192, \pm 0.520)$ .
Graphing calculator required. The polar function $f$ is given by $f (θ) = 3 \cos (1.6 θ) - 0.5$ . Which statement compares the distances from the origin of the points on the polar graph of $r = f (θ)$ at $θ = 0.4$ and at $θ = 2.2$ , with a correct reason?
Answer: D.
1. A
  A negative average rate of change says the signed radius went down, from about 1.906 to about $- 3.288$ . Distance from the origin is $| r |$ , and 3.288 is more than 1.906, so the point ends farther away even though $r$ decreased.
2. B
  Comparing signed radii does not compare distances. At $θ = 2.2$ , $r \approx - 3.288$ puts the point 3.288 units from the origin, on the ray opposite the one for $θ = 2.2$ . That is farther out than the point at $θ = 0.4$ , which is 1.906 units away.
3. C
  The conclusion is right but the reason is not. $f (0.4) \approx 1.906$ is positive, $f$ is 0 at $θ \approx 0.877$ , and $f$ reaches its least value, $- 3.5$ , at $θ \approx 1.963$ and increases after that. So $f$ is neither negative nor decreasing on the whole interval.
4. D
  Correct. The distance from the origin is $| f (θ) |$ . Here $f (0.4) = 3 \cos 0.64 - 0.5 \approx 1.906$ and $f (2.2) = 3 \cos 3.52 - 0.5 \approx - 3.288$ , so the distances are about 1.906 and 3.288, and the point at $θ = 2.2$ is about 1.381 units farther from the origin.

Section II Part A: Free Response (graphing calculator required)

2 questions · 35 minutes · Graphing calculator required

Question 1: Function Concepts 6 points

Let $f$ be an increasing function defined for $0 \leq x \leq 8$ . The table gives values of $f (x)$ at selected values of $x$ . The function $g$ is given by $g (x) = \frac{2 x^{2} + 7 x - 3}{x^{3} + 2}$ .

Selected values of $f (x)$
$x$ 0 2 4 6 8

$f (x)$ $- 7$ 9 21 29 33

A graphing calculator is required for this question. Use it to evaluate $g$ , to find its zeros, and to look at its graph or a table of its values. Give any decimal answer accurate to three places after the decimal point. Respond to parts A, B, and C, labeling subparts i and ii.
1. A.
  
  i. Find the value of $(f \circ g) (1)$ , or indicate that it is not defined.
  
  ii. Find the value of $(g \circ f^{- 1}) (29)$ as a decimal approximation, or indicate that it is not defined.
  Question 1, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Finds $(f \circ g) (1) = f (g (1)) = f (2) = 9$ . Here $g (1) = \frac{2 + 7 - 3}{1 + 2} = 2$ , which is in the domain of $f$ and in the table.Earns it: " $g (1) = 2$ , so $(f \circ g) (1) = f (2) = 9$ "; "9" with $g (1) = 2$ shownDoes not: " $g (f (1))$ " or "not defined because $f (1)$ is not in the table" (the order is reversed: $g$ acts first, then $f$ ); " $f (1) \cdot g (1)$ " (a product, not a composition); "2" (that is $g (1)$ ; $f$ still has to act on it)
  - 1 point Finds $(g \circ f^{- 1}) (29) = g (f^{- 1} (29)) = g (6) = \frac{111}{218} \approx 0.509$ . Because $f$ is increasing it is one-to-one, so $f^{- 1}$ exists, and $f (6) = 29$ gives $f^{- 1} (29) = 6$ . The exact value $\frac{111}{218}$ also earns the point.Earns it: " $f (6) = 29$ , so $f^{- 1} (29) = 6$ and $g (6) = 0.509$ "; " $g (6) = \frac{111}{218} \approx 0.509$ " with $f^{- 1} (29) = 6$ shownDoes not: " $g (29) \approx 0.077$ " (skips the inverse: 29 is an output of $f$ , so $f^{- 1}$ turns it back into the input 6 first); " $\frac{1}{g (6)} \approx 1.964$ " (reads the $- 1$ as a reciprocal); "not defined, because 29 is not an input in the table" (29 is an output of $f$ , which makes it an input of $f^{- 1}$ )
  Model response
  i. $g (1) = \frac{2 (1)^{2} + 7 (1) - 3}{1^{3} + 2} = \frac{6}{3} = 2$ . Since 2 is in the domain of $f$ , $(f \circ g) (1) = f (2) = 9$ .
  
  ii. $f$ is increasing, so each output comes from exactly one input and $f^{- 1}$ exists. The table gives $f (6) = 29$ , so $f^{- 1} (29) = 6$ . Then $(g \circ f^{- 1}) (29) = g (6) = \frac{2 (36) + 7 (6) - 3}{216 + 2} = \frac{111}{218} \approx 0.509.$
2. B.
  
  i. Find all real zeros of $g$ . Express each zero as a decimal approximation.
  
  ii. Determine the end behavior of $g$ as $x$ decreases without bound. Express your answer using the mathematical notation of a limit.
  Question 1, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Finds both real zeros, $x \approx - 3.886$ and $x \approx 0.386$ , and no others. The zeros of $g$ are the zeros of the numerator $2 x^{2} + 7 x - 3$ , since neither is a zero of the denominator, whose only real zero is $x = - \sqrt[3]{2} \approx - 1.260$ . The exact values $\frac{- 7 \pm \sqrt{73}}{4}$ also earn the point.Earns it: " $x = - 3.886$ and $x = 0.386$ " from a graph or a solver; " $2 x^{2} + 7 x - 3 = 0$ , so $x = \frac{- 7 \pm \sqrt{73}}{4}$ , about $- 3.886$ and $0.386$ "Does not: only one of the two zeros; " $x = - 1.260$ " (that is where the denominator is 0, a vertical asymptote, not a zero); " $g (0) = - \frac{3}{2}$ " (that is the vertical intercept, not a zero)
  - 1 point States $\lim_{x \to - \infty} g (x) = 0$ . The numerator has degree 2 and the denominator degree 3, so the denominator grows faster in size and the outputs approach 0.Earns it: " $\lim_{x \to - \infty} g (x) = 0$ "; " $\lim_{x \to - \infty} \frac{2 x^{2} + 7 x - 3}{x^{3} + 2} = 0$ , since the degree of the numerator is less than the degree of the denominator"Does not: " $y = 0$ " or " $g$ approaches 0" (correct value, but not in limit notation); " $\lim_{x \to - \infty} g (x) = 2$ " (the ratio of the leading coefficients is the limit only when the degrees are equal); " $\lim_{x \to - \infty} g (x) = - \infty$ " (the denominator has the higher degree, so the outputs shrink toward 0)
  Model response
  i. The denominator $x^{3} + 2$ is 0 only at $x = - \sqrt[3]{2} \approx - 1.260$ , where the numerator is not 0, so $g (x) = 0$ exactly where $2 x^{2} + 7 x - 3 = 0$ . Graphing $g$ and using the zero feature gives $x \approx - 3.886$ and $x \approx 0.386$ .
  
  ii. The numerator has degree 2 and the denominator has degree 3. As $x$ decreases without bound, the denominator grows faster in size than the numerator, so the quotient gets closer and closer to 0. $\lim_{x \to - \infty} g (x) = 0$ .
3. C.
  
  i. Of the function types linear, quadratic, and exponential, which best models $f$ for $0 \leq x \leq 8$ ?
  
  ii. Give a reason for your answer in part C (i) based on how the output values of $f$ change over the equal-length input-value intervals in the table. Refer to values in the table in your reasoning.
  Question 1, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Identifies a quadratic function. The point is for the function type alone; an expression is not asked for and is not scored.Earns it: "Quadratic"; "Quadratic, $f (x) = - 0.5 x^{2} + 9 x - 7$ "Does not: "Linear" (the first differences 16, 12, 8 and 4 are not constant); "Exponential" (the ratios of consecutive outputs, $\frac{9}{- 7}$ , $\frac{21}{9}$ , $\frac{29}{21}$ and $\frac{33}{29}$ , are not constant, and the outputs even change sign)
  - 1 point Gives a reason based on constant second differences: over consecutive input intervals of length 2, the first differences in the outputs are $9 - (- 7) = 16$ , $21 - 9 = 12$ , $29 - 21 = 8$ and $33 - 29 = 4$ , and the second differences are all $- 4$ . Saying that the first differences decrease by the same amount, 4, each time is the same reason and earns the point.Earns it: "The first differences are 16, 12, 8, 4 and the second differences are $- 4$ , $- 4$ , $- 4$ , constant over intervals of length 2."; "Each time $x$ goes up by 2, the increase in $f (x)$ is 4 less than the time before: 16, 12, 8, 4."Does not: "The first differences 16, 12, 8, 4 are not constant, so it is not linear" alone (rules out linear but does not show the change in the differences is constant); " $f$ increases more and more slowly" (an exponential or logarithmic function could do that too); "The rate of change is decreasing" with no values from the table
  Model response
  i. Quadratic.
  
  ii. The inputs go up by 2 each step. The first differences in the outputs are $9 - (- 7) = 16$ , $21 - 9 = 12$ , $29 - 21 = 8$ and $33 - 29 = 4$ , and the second differences are $12 - 16 = - 4$ , $8 - 12 = - 4$ and $4 - 8 = - 4$ . The second differences are a constant $- 4$ over consecutive equal-length input intervals, so a quadratic model fits best. For example, $f (x) = - 0.5 x^{2} + 9 x - 7$ matches all five values.
Your score: 0 of 6, from the boxes you ticked.
Question 2: Modeling a Non-Periodic Context 6 points

After a heavy rainstorm, a park ranger measures the water level of a creek at a marked post. The rain stopped at 6 p.m., at time $t = 0$ hours. At $t = 2$ , the water stood 46 centimeters above the creek's normal level. At $t = 5$ it stood 78 centimeters above normal, and at $t = 10$ it stood 61 centimeters above normal. The ranger models the water level, in centimeters above normal, $t$ hours after the rain stopped by the function $L$ given by $L (t) = a t^{2} + b t + c$ , where $a$ , $b$ , and $c$ are constants. The model starts at $t = 0$ and assumes that the level follows one quadratic pattern through all three measurements.

A graphing calculator is required for this question. Use it to solve for the constants, to evaluate $L$ , and to solve equations. Store values instead of rounding along the way, and give decimal answers accurate to three places after the decimal point. Respond to parts A, B, and C, labeling subparts i and ii.
1. A.
  
  i. Use the given data to write three equations that can be used to find the values of the constants $a$ , $b$ , and $c$ in the expression for $L (t)$ .
  
  ii. Find the values of $a$ , $b$ , and $c$ as decimal approximations.
  Question 2, part A: your response. It stays in this browser tab and is never sent anywhere.
  Score part A
  - 1 point Writes the three equations $4 a + 2 b + c = 46$ , $25 a + 5 b + c = 78$ , and $100 a + 10 b + c = 61$ , from $L (2) = 46$ , $L (5) = 78$ , and $L (10) = 61$ .Earns it: " $a (2)^{2} + b (2) + c = 46$ , $a (5)^{2} + b (5) + c = 78$ , $a (10)^{2} + b (10) + c = 61$ "; " $4 a + 2 b + c = 46$ , $25 a + 5 b + c = 78$ , $100 a + 10 b + c = 61$ "Does not: only two of the three equations (three unknown constants need three equations); " $2 a + 2 b + c = 46$ " and similar (the input is squared in the first term: $a (2)^{2} = 4 a$ ); " $46 a^{2} + 46 b + c = 2$ " and similar (puts the level in for $t$ and the time in for $L (t)$ )
  - 1 point Finds $a \approx - 1.758$ , $b = 22.975$ , and $c \approx 7.083$ . The exact values are $a = - \frac{211}{120}$ , $b = \frac{919}{40}$ , and $c = \frac{85}{12}$ , and they also earn the point. Any method earns the point: solving the system by elimination or with matrices, or a quadratic regression on the three points, which fits them exactly.Earns it: " $a = - 1.758$ , $b = 22.975$ , $c = 7.083$ " from a matrix or a solver; " $L (t) = - 1.758 t^{2} + 22.975 t + 7.083$ " from a quadratic regression on the three pointsDoes not: " $a = 1.758$ " (a positive $a$ makes the parabola open up, and then the level could not rise to 78 and fall back to 61); values for only two of the constants
  Model response
  i. $L (2) = 46$ , $L (5) = 78$ , and $L (10) = 61$ give $4 a + 2 b + c = 46, 25 a + 5 b + c = 78, 100 a + 10 b + c = 61.$
  
  ii. Solving the system with a matrix on the calculator gives $a \approx - 1.758$ , $b = 22.975$ , and $c \approx 7.083$ , so $L (t) \approx - 1.758 t^{2} + 22.975 t + 7.083$ , with the unrounded values stored.
2. B.
  
  i. Use model $L$ to find the average rate of change of the water level, in centimeters per hour, from $t = 7$ to $t = 11$ . Show the computations that lead to your answer.
  
  ii. Interpret the meaning of your answer from part B (i) in the context of the problem.
  Question 2, part B: your response. It stays in this browser tab and is never sent anywhere.
  Score part B
  - 1 point Computes $\frac{L (11) - L (7)}{11 - 7} = \frac{47.050 - 81.750}{4} = - 8.675$ centimeters per hour, with the computation shown.Earns it: " $\frac{L (11) - L (7)}{4} = \frac{47.05 - 81.75}{4} = - 8.675$ "Does not: " $- 8.669$ " (comes from rounding $a$ , $b$ , and $c$ to three places before computing); " $L (11) - L (7) = - 34.7$ " (a change in level, not a rate); " $\frac{61 - 78}{10 - 5} = - 3.4$ " (the average rate between two measurements, not over 7 to 11 hours with the model)
  - 1 point Interprets the value in context with units and the time interval: from 7 to 11 hours after the rain stopped, the water level fell by an average of 8.675 centimeters per hour. Saying it changed by $- 8.675$ centimeters per hour, on average, over that interval also earns the point.Earns it: "Between 7 and 11 hours after the rain stopped, the creek's level dropped by about 8.675 centimeters per hour on average."Does not: "The level drops 8.675 centimeters every hour" (no time interval, and treats the rate as constant); "The level is 8.675 centimeters lower" (a change, not a rate, and no interval); "The level rises at 8.675 centimeters per hour" (the rate is negative, so the level is falling)
  Model response
  i. With the stored constants, $L (7) = 81.750$ and $L (11) = 47.050$ , so $\frac{L (11) - L (7)}{11 - 7} = \frac{47.050 - 81.750}{4} = - 8.675 centimeters per hour.$
  
  ii. From 7 hours to 11 hours after the rain stopped, the water level in the creek went down by an average of 8.675 centimeters per hour.
3. C.
  
  Use model $L$ to respond to the following.
  
  i. The model predicts that the level rises to a crest and then falls. Write and solve an equation whose solution gives the time $t$ , after the crest, when the model predicts a water level of 30 centimeters above normal. Express your value for $t$ as a decimal approximation.
  
  ii. The ranger's records for this creek show that after a storm, the water usually falls back much more slowly than it rose. If that happens this time, will the level actually come back down to 30 centimeters above normal before or after the time you found in part C (i)? Explain your reasoning using a property of the graph of $L$ .
  Question 2, part C: your response. It stays in this browser tab and is never sent anywhere.
  Score part C
  - 1 point Writes $L (t) = 30$ , that is, $a t^{2} + b t + c = 30$ with the values from part A, and gives the solution after the crest, $t \approx 11.978$ hours. The equation has a second solution, $t \approx 1.088$ , which is before the crest at $t \approx 6.533$ and is not the time asked for.Earns it: " $- 1.758 t^{2} + 22.975 t + 7.083 = 30$ , $t = 11.978$ " with stored values; " $L (t) = 30$ at $t = 1.088$ and $t = 11.978$ ; after the crest, $t = 11.978$ "Does not: " $t = 11.978$ " with no equation; " $t = 1.088$ " (that is when the level passed 30 centimeters on the way up); " $t = 11.981$ " (comes from rounding $a$ , $b$ , and $c$ before solving)
  - 1 point Answers after, and explains with a property of the quadratic model: the graph of $L$ is a parabola, symmetric about the vertical line through its crest, $t \approx 6.533$ . So the model has the level fall from the crest to 30 centimeters in the same time it took to rise from 30 centimeters to the crest, about 5.445 hours each way, from $t \approx 1.088$ to $t \approx 6.533$ and on to $t \approx 11.978$ . If the water falls more slowly than it rose, it takes longer than that to get back down to 30 centimeters, so the actual time is later than $t \approx 11.978$ . Two other explanations also earn the point. One says the model's rates of change after the crest mirror its rates before the crest, so a slower fall means a later time. The other uses concavity: $L$ is concave down, so after the crest its rate of change keeps getting more negative and the model's fall speeds up, while a creek that falls more slowly than it rose takes longer to get down to 30 centimeters.Earns it: "After. A parabola is symmetric about its vertex, so $L$ assumes the creek falls as fast as it rose. A slower fall means it reaches 30 cm later than 11.978 hours."; "After, because $L$ predicts 30 cm at 1.088 hours and at 11.978 hours, the same 5.445 hours before and after the crest at 6.533. A slower fall takes more than 5.445 hours."; "After. $L$ is concave down, so after the crest its rate of change keeps getting more negative and the model's fall speeds up. A creek that falls more slowly than it rose takes longer, so it reaches 30 cm after $t = 11.978$ ."Does not: "After, because creeks go down slowly" (the right answer with no property of model $L$ ); "Before, because the level is falling more slowly" (a slower fall keeps the water high for longer); "After, because the model eventually gives negative values" (true of $L$ , but it does not explain why the slower fall changes the time)
  Model response
  i. $L (t) = 30$ , that is, $a t^{2} + b t + c = 30$ with the stored values of $a$ , $b$ , and $c$ . Graphing $y = L (t)$ and $y = 30$ shows two intersections, at $t \approx 1.088$ and $t \approx 11.978$ . The crest is at $t = - \frac{b}{2 a} \approx 6.533$ , so the time after the crest is $t \approx 11.978$ hours.
  
  ii. After. The graph of $L$ is a parabola, which is symmetric about the vertical line through its vertex, $t \approx 6.533$ . So the model has the creek go down at exactly the pace it came up: it passes 30 centimeters about 5.445 hours before the crest, at $t \approx 1.088$ , and again about 5.445 hours after it, at $t \approx 11.978$ . If this time the water falls more slowly than it rose, it needs more than 5.445 hours after the crest to drop to 30 centimeters, so it gets there later than $t \approx 11.978$ . The model's answer is too early.
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