Unit 1 check

Unit 1 check: Polynomial and Rational Functions

14 questions from 14 lessons in Unit 1. Answer each one, and if you miss, follow the link back to where it is taught, then try again. Your score counts first tries.

Question 1 · from 1.1 · Match

No calculator. A garage charges $C (h)$ dollars to park for $h$ hours. Tap a term, then the phrase it matches.

Question 2 · from 1.2 · Sort it

No calculator. Tap a situation, then tap the sign of the rate of change of the second quantity with respect to the first.

Positive

Negative

Question 3 · from 1.3 · Calculate

No calculator. A quadratic function $p$ has average rates of change 5, 2 and $- 1$ over the intervals $[0, 1]$ , $[1, 2]$ and $[2, 3]$ . What is its average rate of change over $[3, 4]$ ?

Your answer

Question 4 · from 1.4 · Multiple choice

No calculator. Which description fits a point of inflection of a polynomial function?

Setup for question 5

A small company models its monthly profit, in thousands of dollars, from making $x$ hundred units of a product: $P (x) = - 0.5 x (x - 2) (x - 9)$ for $0 < x \leq 10$ . Below 200 units it cannot cover its costs, and above 900 it has to pay for overtime.

Testing the sign of $P$ between its zeros
Interval	Test value	Value of $P$
$0 < x < 2$	1	−4, negative
$2 < x < 9$	5	30, positive
$9 < x \leq 10$	10	−40, negative

Question 5 · from 1.5 · Multiple choice

No calculator. For which production levels in the domain is $P (x) \geq 0$ ?

Setup for question 6

Let $p (x) = 2 x^{3} - 50 x^{2} + 7$ . Going left, $p (- 10) = - 6,993$ , $p (- 100) = - 2,499,993$ and $p (- 1000) = - 2,049,999,993$ . Going right, $p (10) = - 2,993$ , $p (100) = 1,500,007$ and $p (1000) = 1,950,000,007$ . The table gives the ratio of $p (x)$ to its leading term, $\frac{p (x)}{2 x^{3}}$ , to four decimal places.

A polynomial against its leading term
$x$	$\frac{p (x)}{2 x^{3}}$
−1000	1.0250
−100	1.2500
−10	3.4965
10	−1.4965
100	0.7500
1000	0.9750

Question 6 · from 1.6 · Multiple choice

No calculator. Which statement describes the left end of $p$ ?

Question 7 · from 1.7 · Spot the error

No calculator. A student finds the horizontal asymptote of $r (x) = \frac{1 + 4 x^{3}}{2 x^{3} - x}$ . Tap the first line with a mistake.

Question 8 · from 1.8 · Fill the statement

No calculator. Let $r (x) = \frac{x (x + 6)}{(x + 6) (x - 2)}$ . Tap a blank, then tap an answer.

Of the two zeros of the numerator, also makes the denominator zero, so the only real zero of $r$ is .

Question 9 · from 1.9 · Calculate

No calculator. The graph of $r (x) = \frac{x + 4}{x^{2} + x - 12}$ has exactly one vertical asymptote, $x = k$ . What is $k$ ?

Your answer

Question 10 · from 1.10 · Calculate

No calculator. The graph of $r (x) = \frac{x^{2} - x - 6}{x - 3}$ has a hole at $x = 3$ . What is the $y$ -coordinate of the hole?

Your answer

Question 11 · from 1.11 · Calculate

No calculator. Divide $x^{2} + 5 x + 9$ by $x + 2$ using long division. What is the remainder?

Your answer

Setup for question 12

Selected values of a function $f$ are below. Let $g (x) = 2 f (x + 1) - 3$ .

Values of $f$
$x$	$f (x)$
−2	3
0	−1
2	5
4	0

Question 12 · from 1.12 · Multiple choice

No calculator. The point $(4, 0)$ is on the graph of $f$ . Which point is its image on the graph of $g$ ?

Question 13 · from 1.13 · Sort it

No calculator. A school models ticket sales for its play. Tap each statement, then what kind of statement it is.

Assumption

Domain restriction

Range restriction

Setup for question 14

A museum counts weekly visitors, in hundreds, $w$ weeks after a new exhibit opens: 5.0, 16.2, 25.8, 33.8 and 40.2 for $w = 0$ to 4. The first differences, 11.2, 9.6, 8.0 and 6.4, fall by 1.6 each week, so a quadratic regression is the right one to run.

Weekly visitors, in hundreds
$w$ (weeks)	$V (w)$
0	5.0
1	16.2
2	25.8
3	33.8
4	40.2

Question 14 · from 1.14 · Calculate

Graphing calculator required. A museum counts weekly visitors, in hundreds, $w$ weeks after a new exhibit opens: 5.0, 16.2, 25.8, 33.8 and 40.2 for $w = 0, 1, 2, 3, 4$ . Run a quadratic regression on these data, and use it to predict the number of visitors, in hundreds, in week 6.

Your answer hundred visitors

0 of 14 answered.