Unit 3 · Topic 3.8 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Precalculus 3.8: The Tangent Function

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

The graph of y = tan theta for theta from -5 to 5 radians, drawn as separate branches with dashed vertical asymptotes at theta = -3pi/2, -pi/2, pi/2 and 3pi/2, where cos theta = 0. No branch crosses an asymptote. Each full branch rises from left to right, concave down on its left half and concave up on its right half, with a point of inflection on the theta-axis at theta = -pi, 0 and pi. A green bar along the axis from 0 to pi is labeled period pi.

Part 1 Think first Water on the fence

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A lawn sprinkler has one straight arm with a nozzle at each end, spraying in opposite directions. The arm spins at a steady rate, one full turn every 8 seconds. A long straight fence runs past it, 2 meters away. The spot where the water hits the fence slides along, runs off far down the fence, then comes back from the other end. How often does that pattern repeat?

In short. The tangent function gives the slope of an angle's terminal ray, $\tan θ = \frac{\sin θ}{\cos θ}$ , and its graph repeats every $π$ with a vertical asymptote wherever $\cos θ = 0$ .

Half a turn points the terminal ray the other way along the same line, so the slope repeats and the period of $\tan θ$ is $π$ , not $2 π$ .
The graph has vertical asymptotes at $θ = \frac{π}{2} + k π$ , where $\cos θ = 0$ . Between consecutive asymptotes it increases, concave down and then concave up.
In $f (θ) = a \tan (b (θ + c)) + d$ , $| a |$ is a vertical dilation, the period is $| \frac{1}{b} | π$ , the phase shift is $- c$ , and the points of inflection lie on $y = d$ . There is no amplitude.

Part 2 Learn it Four short ideas, each checked as you go

2a. Tangent is the slope of the terminal ray

Put an angle $θ$ in standard position. Its terminal ray meets the unit circle at $P = (\cos θ, \sin θ)$ , and the tangent function gives the slope of that ray. The ray starts at the origin, so its slope is rise over run from $(0, 0)$ to $P$ :

\tan θ = \frac{\sin θ - 0}{\cos θ - 0} = \frac{\sin θ}{\cos θ}, \cos θ \neq 0

Now add $π$ , half a turn. The new terminal ray points the opposite way along the same line through the origin, and $P$ moves to $(- \cos θ, - \sin θ)$ . Both coordinates change sign, so their ratio does not:

\tan (θ + π) = \frac{- \sin θ}{- \cos θ} = \tan θ

Sine and cosine need a full turn, $2 π$ , before they repeat. The slope repeats every half turn, so the period of the tangent function is $π$ .

Here is one pair. $\tan \frac{π}{4} = 1$ , because $\sin \frac{π}{4} = \cos \frac{π}{4}$ . The ray for $\frac{5 π}{4}$ points into Quadrant III, but it lies on the same line $y = x$ , so $\tan \frac{5 π}{4} = 1$ as well.

Quick check · Multiple choice

No calculator. The terminal ray of an angle $θ$ in standard position meets the unit circle at $P = (0.8, - 0.6)$ . What is $\tan (θ + π)$ ?

2b. Asymptotes where cosine is zero

The quotient $\frac{\sin θ}{\cos θ}$ cannot be evaluated where $\cos θ = 0$ . That happens at $\frac{π}{2}$ , at $\frac{3 π}{2}$ , at $- \frac{π}{2}$ , and in general at $θ = \frac{π}{2} + k π$ for every integer $k$ . The terminal ray there is vertical, and a vertical line has no slope.

Watch the slope on the way there. As $θ$ climbs toward $\frac{π}{2}$ the ray gets steeper and steeper: $\tan 1.5 \approx 14$ and $\tan 1.57 \approx 1256$ . Just past $\frac{π}{2}$ the ray tips into Quadrant II and the slope is a huge negative number. So the graph has a vertical asymptote at every $\frac{π}{2} + k π$ . That is the sprinkler's water running off one end of the fence and coming back from the other.

Where $\sin θ = 0$ , at $θ = k π$ , the ray is horizontal and $\tan θ = 0$ . Those are zeros of tangent, not asymptotes.

Between two consecutive asymptotes, say $- \frac{π}{2}$ and $\frac{π}{2}$ , the tangent increases the whole way. Its graph is concave down on the left half and concave up on the right half. It changes concavity at the point of inflection halfway between, $(0, 0)$ .

Quick check · Fill the statement

No calculator. Tap a blank, then tap an answer.

The graph of $y = \tan θ$ repeats every radians. It has vertical asymptotes at $θ =$ for every integer $k$ , because at those inputs. Between two consecutive asymptotes the graph increases and changes from concave to concave .

2c. Stretching and lifting: a and d

Multiply by $a$ and every output of $\tan θ$ gets multiplied by $a$ . The graph of $g (θ) = a \tan θ$ is a vertical dilation of $y = \tan θ$ by a factor of $| a |$ . At $θ = \frac{π}{4}$ , $\tan θ = 1$ , so $3 \tan θ$ is 3 there: the same shape, only steeper. The zeros, the asymptotes and the period stay put. Three times 0 is still 0, and multiplying cannot rescue a slope that does not exist.

If $a < 0$ , the dilation comes with a reflection over the $x$ -axis. The graph of $- 2 \tan θ$ decreases between its asymptotes instead of increasing.

Add $d$ and the whole graph moves up $d$ units. For $g (θ) = \tan θ + 2$ , the points of inflection move from the $θ$ -axis up to $y = 2$ , the line containing the points of inflection.

That line is tempting to call a midline, and $| a |$ is tempting to call an amplitude. Both are mistakes. A sinusoid's midline sits halfway between its maximum and minimum, and its amplitude is half the distance between them. Tangent takes every real value, so it has no maximum, no minimum and no amplitude. The number $| a |$ tells you how steep the graph is, not how high it goes.

Quick check · Decide and justify

No calculator. A classmate says the graph of $g (θ) = 3 \tan θ + 2$ has an amplitude of 3 and a midline of $y = 2$ . Is the classmate right?

Which reason supports your answer?

2d. Squeezing and sliding: b and c

A factor $b$ inside the tangent changes how fast the angle turns. The graph of $\tan (b θ)$ has asymptotes where $b θ = \frac{π}{2} + k π$ , so they sit $| \frac{1}{b} |$ times as far apart as the asymptotes of $\tan θ$ . The period of $\tan (b θ)$ is $| \frac{1}{b} | π$ .

So $\tan (2 θ)$ has period $\frac{π}{2}$ , and $\tan (\frac{θ}{3})$ has period $3 π$ . If you remember sine, you will want to write $\frac{2 π}{| b |}$ . For tangent that is twice too long, because tangent starts from a period of $π$ . If $b < 0$ , the graph is also reflected over the $y$ -axis.

Adding $c$ inside slides the graph sideways. The graph of $\tan (θ + c)$ is a phase shift of $- c$ , so $\tan (θ - \frac{π}{4})$ slides right by $\frac{π}{4}$ and its point of inflection moves from $(0, 0)$ to $(\frac{π}{4}, 0)$ .

Put together, the graph of $f (θ) = a \tan (b (θ + c)) + d$ has

a vertical dilation by $| a |$ , reflected over the $x$ -axis if $a < 0$ ,
period $| \frac{1}{b} | π$ and phase shift $- c$ ,
its points of inflection on the line $y = d$ .

One point of inflection is $(- c, d)$ , and the asymptotes sit half a period to either side of it.

Quick check · Sort it

No calculator. Each card is a transformation of $y = \tan θ$ . Tap a card, then the period of its graph.

Period $\frac{π}{2}$

Period $π$

Period $2 π$

Part 3 Watch it done Pin down one transformed tangent

Watch each step, then do the last one yourself.

Let $g (x) = - 2 \tan (\frac{π}{6} (x - 1)) + 4$ . Find the first vertical asymptote of the graph of g to the right of $x = 0$ .

Step 1 · Read the parameters
Match g to $a \tan (b (x + c)) + d$ . Here $a = - 2$ , $b = \frac{π}{6}$ , $c = - 1$ and $d = 4$ . Read $c$ with care: $x - 1$ is $x + (- 1)$ .
Step 2 · Find the period
The period is $| \frac{1}{b} | π = \frac{6}{π} \cdot π = 6$ . Consecutive asymptotes of g are 6 apart.
Step 3 · Place a point of inflection
At $x = 1$ the output is $- 2 \tan 0 + 4 = 4$ , so $(1, 4)$ is a point of inflection, on the line $y = 4$ . Because $a < 0$ , g decreases between its asymptotes. The asymptotes sit half a period to either side of each point of inflection.
Why this step?
Why is there a point of inflection at $x = 1$ , and not at $x = - 1$ ?
The phase shift is $- c = 1$ . Another way to see it: $\tan u$ has a point of inflection at $u = 0$ , and the inside, $\frac{π}{6} (x - 1)$ , is 0 when $x = 1$ . The sign inside the parentheses is the opposite of the direction of the shift.
Your step · Calculate

No calculator. Finish it. What is the $x$ -value of the first vertical asymptote of the graph of $g (x) = - 2 \tan (\frac{π}{6} (x - 1)) + 4$ to the right of $x = 0$ ?

Your answer

Part 4 Try it Move the asymptotes yourself

The grapher draws $g (x) = a \tan (b (x + c)) + d$ in color over the dashed parent $y = \tan x$ . Its asymptotes are dashed vertical lines and the line containing its points of inflection is a solid gray horizontal line. Use it to test the rules from Learn it before you trust them on paper.

Lab

Function Grapher

Start with $b = 2$ and leave $a = 1$ , $c = 0$ and $d = 0$ . Before you look, say how far apart the asymptotes will be, then check against the panel. Next set $c = 1$ and $d = - 1$ and predict where the point of inflection that was at the origin lands. Then set $a = - 1.5$ and say what changes and what stays put. Work the question below on paper.

Open the full Function Grapher lab

Try it · Multiple choice

No calculator. Let $k (x) = \tan (\frac{1}{2} (x - 1)) + 2$ . What is the $x$ -value of the first vertical asymptote of the graph of k to the right of $x = 1$ ?

Part 5 Explain it Write it the way the exam asks

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 37.5% of the exam score. This one is shaped like Question 4, Symbolic Manipulations.

Free-response practice · Question 4 style

No calculator. Let $f (θ) = 3 \tan (2 θ) - 1$ . (A) Find the exact value of $f (\frac{π}{8})$ . Show the work that leads to your answer. (B) Rewrite $f (θ)$ as an expression in $\sin (2 θ)$ and $\cos (2 θ)$ . Use it to find every value of $θ$ at which f is not defined. (C) Find the period of f and an equation of the line containing the points of inflection of the graph of f.

Your answer stays on this page. It is never sent or saved.

Model answer

(A) $f (\frac{π}{8}) = 3 \tan \frac{π}{4} - 1$ . Since $\sin \frac{π}{4} = \cos \frac{π}{4} = \frac{\sqrt{2}}{2}$ , $\tan \frac{π}{4} = 1$ , so $f (\frac{π}{8}) = 3 - 1 = 2$ . (B) $f (θ) = \frac{3 \sin (2 θ)}{\cos (2 θ)} - 1$ . This is not defined where $\cos (2 θ) = 0$ , which is where $2 θ = \frac{π}{2} + k π$ , so $θ = \frac{π}{4} + \frac{k π}{2}$ for every integer $k$ . (C) The period is $| \frac{1}{2} | π = \frac{π}{2}$ . The vertical translation is $- 1$ , so the points of inflection lie on the line $y = - 1$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Uses $\tan \frac{π}{4} = 1$ to find $f (\frac{π}{8}) = 2$ , with the work shown.
Writes $f (θ) = \frac{3 \sin (2 θ)}{\cos (2 θ)} - 1$ .
Finds $θ = \frac{π}{4} + \frac{k π}{2}$ from $\cos (2 θ) = 0$ , dividing every term by 2.
Gives the period $\frac{π}{2}$ and the line $y = - 1$ .

Part 6 Check yourself A table of a transformed tangent

The three case questions use the same function h.

The function h is given by $h (x) = a \tan (b x) + d$ for every $x$ where $\tan (b x)$ is defined, where $a$ , $b$ and $d$ are constants with $a > 0$ and $b > 0$ . The graph of h has vertical asymptotes at $x = - 2$ and $x = 2$ and no vertical asymptote between them. The table gives values of h, rounded to two decimal places where they are not whole numbers.

Values of the transformed tangent function $h$
$x$	$- 1.5$	$- 1$	$- 0.5$	$0$	$0.5$	$1$	$1.5$
$h (x)$	$- 3.83$	$- 1$	$0.17$	$1$	$1.83$	$3$	$5.83$

Case question 1 · Multiple choice

No calculator. Which equation defines h?

Case question 2 · Calculate

No calculator. What is $h (3)$ ?

Your answer

Case question 3 · Calculate

Graphing calculator required. What is $h (1.8)$ ? Give a decimal to two places.

Your answer

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the period and the asymptotes of $g (θ) = 4 \tan (\frac{θ}{2}) - 1$ . Tap the first line that goes wrong.

Check yourself · Match

No calculator. Each description is one branch of a graph: the two vertical asymptotes on either side of it, its point of inflection, and whether it rises or falls from left to right. Tap an equation, then its branch.

Part 7 Keep it Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 3.5 Sinusoidal Functions · Sort it

No calculator. Each card names a function and an open interval of $θ$ in radians. Tap a card, then how the graph behaves on that whole interval.

Increasing, concave down

Decreasing, concave down

Decreasing, concave up

Increasing, concave up

From 3.7 Sinusoidal Function Context and Data Modeling · Calculate

No calculator. On a school day, the temperature in a classroom is modeled by $T (h) = 4 \sin (\frac{π}{12} (h - 3)) + 20$ degrees Celsius, where $h$ is the number of hours after 6 a.m., for $0 \leq h \leq 12$ . What temperature does the model predict at 11 a.m.?

Your answer degrees Celsius

In three lines

$\tan θ = \frac{\sin θ}{\cos θ}$ is the slope of the terminal ray, and the slope repeats every half turn, so the period is $π$ .
Vertical asymptotes sit at $θ = \frac{π}{2} + k π$ , where $\cos θ = 0$ . Between them tangent increases, concave down and then concave up.
$a \tan (b (θ + c)) + d$ has vertical dilation $| a |$ , period $| \frac{1}{b} | π$ , phase shift $- c$ , points of inflection on $y = d$ , and no amplitude.

Vocabulary

Tangent function: The function $f (θ) = \tan θ$ , which gives the slope of the terminal ray of $θ$ and equals $\frac{\sin θ}{\cos θ}$ where $\cos θ \neq 0$ .
Period: The tangent function has a period of $π$ , because the slope values of the terminal ray repeat every one-half revolution of the circle.
Points of inflection: Where the graph of $y = \tan θ$ changes from concave down to concave up, halfway between consecutive asymptotes; for $y = \tan θ + d$ they lie on a line moved by $d$ units.
Vertical dilation: The graph of $g (θ) = a \tan θ$ is a vertical dilation of the graph of $y = \tan θ$ by a factor of $| a |$ , with a reflection over the $x$ -axis if $a < 0$ .
Horizontal dilation: The graph of $g (θ) = \tan (b θ)$ is a horizontal dilation of the graph of $y = \tan θ$ whose period differs by a factor of $| \frac{1}{b} |$ .
Phase shift: The graph of $g (θ) = \tan (θ + c)$ is a horizontal translation of the graph of $y = \tan θ$ by $- c$ units.

Course alignment, for teachers

AP Precalculus topic 3.8, Unit 3: Trigonometric and Polar Functions.

Skill 2.A: Identify information from graphical, numerical, analytical, and verbal representations to answer a question or construct a model, with and without technology.
Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.

This lesson maps all 10 essential knowledge statements of the topic to the part that teaches each one: 10 are checked by a scored question. The map is in the lesson file.