AP Precalculus Quiz: The Tangent Function

What this quiz covers

This quiz focuses on The Tangent Function, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. It translates the graph horizontally $\frac{\pi}{3}$ units to the left. (correct answer)
2. It translates the graph horizontally $\frac{\pi}{3}$ units to the right.
3. It translates the graph vertically $\frac{\pi}{3}$ units up.
4. It changes the period of the function to be $\frac{\pi}{3}$ units.

Explanation: For a function of the form $f(x+c)$, the graph is translated horizontally. If $c > 0$, the shift is to the left. Here, $c = \frac{\pi}{3}$, so the graph of $y=\tan(x)$ is shifted $\frac{\pi}{3}$ units to the left.

Question 2

1. Because both $\sin(x+\pi) = -\sin(x)$ and $\cos(x+\pi) = -\cos(x)$, their ratio remains unchanged. (correct answer)
2. Because the period of a ratio of functions is always half the period of the individual functions.
3. Because the tangent function has asymptotes which restrict the period to be smaller than $2\pi$.
4. Because the values of tangent in quadrant I are the reciprocals of values in quadrant IV.

Explanation: The period is the smallest positive value $P$ such that $f(x+P)=f(x)$. Let's test $P=\pi$. $\tan(x+\pi) = \frac{\sin(x+\pi)}{\cos(x+\pi)} = \frac{-\sin(x)}{-\cos(x)} = \frac{\sin(x)}{\cos(x)} = \tan(x)$. Since the function values repeat every $\pi$ units and this is the smallest such positive value, the period is $\pi$.

Question 3

1. $\theta=\sin^{-1}\!\left(\dfrac{1.2}{6.0}\right)$
2. $\theta=\tan^{-1}\!\left(\dfrac{1.2}{6.0}\right)$ (correct answer)
3. $\theta=\cos^{-1}\!\left(\dfrac{1.2}{6.0}\right)$
4. $\theta=\tan^{-1}\!\left(\dfrac{6.0}{1.2}\right)$

Explanation: This question tests AP Precalculus understanding of the tangent function's properties, specifically its application to real-world angle calculations using inverse functions. The tangent function is defined as the ratio of sine to cosine, and when given a tangent value, we use the inverse tangent function to find the angle. In this question, the ramp scenario provides a rise of 1.2 m and a run of 6.0 m, establishing that tan θ = 1.2/6.0. Choice B is correct because it properly applies the inverse tangent function to the given ratio: θ = tan⁻¹(1.2/6.0). Choice D is incorrect because it inverts the ratio to 6.0/1.2, which would give the cotangent rather than the tangent of the angle. To help students: Emphasize that tangent equals rise over run in right triangle applications. Practice setting up ratios correctly before applying inverse functions, and use diagrams to visualize which sides represent rise and run.

Question 4

1. It translates the graph of $y=\tan(\pi x)$ three units up. (correct answer)
2. It translates the graph of $y=\tan(\pi x)$ three units to the right.
3. It vertically stretches the graph of $y=\tan(\pi x)$ by a factor of 3.
4. It changes the period of the function from $\pi$ to $3\pi$.

Explanation: For a function of the form $g(x) = f(x) + d$, the parameter $d$ represents a vertical translation. In this case, $d=3$, so the graph of $y=\tan(\pi x)$ is shifted vertically up by 3 units.

Question 5

1. $b = \frac{1}{2}$ and $c=0$ (correct answer)
2. $b = 2$ and $c = \frac{\pi}{2}$
3. $b = \frac{1}{2}$ and $c=\pi$
4. $b = 2$ and $c=0$

Explanation: The period is given by $\frac{\pi}{|b|}$. If the period is $2\pi$, then $\frac{\pi}{|b|} = 2\pi$, which means $|b| = \frac{1}{2}$. The asymptotes of the parent function are at $\theta = \frac{\pi}{2} + k\pi$. For the transformed function, the asymptotes are at $b(x-c) = \frac{\pi}{2} + k\pi$. If we test $b=\frac{1}{2}$ and $c=0$, we have $\frac{1}{2}x = \frac{\pi}{2} + k\pi$, which simplifies to $x = \pi + 2k\pi$. For $k=0$, we get an asymptote at $x=\pi$, which matches the given information.

Question 6

1. The limit as $x \to \infty$ is $\infty$, and the limit as $x \to -\infty$ is $-\infty$.
2. The limits as $x \to \infty$ and $x \to -\infty$ are both equal to 0.
3. The limits as $x \to \infty$ and $x \to -\infty$ do not exist because the function oscillates without approaching a single value. (correct answer)
4. The limits as $x \to \infty$ and $x \to -\infty$ are both equal to 1.

Explanation: The tangent function is periodic and its values range over all real numbers, $(-\infty, \infty)$, within each period. Because of this periodic oscillation over an unbounded range, the function does not approach a single finite value or consistently grow to infinity. Therefore, the limits do not exist.

Question 7

1. A vertical stretch by a factor of 3 and a horizontal compression by a factor of 2. (correct answer)
2. A horizontal shift of $\frac{\pi}{4}$ units to the right.
3. A vertical shift of 1 unit up.
4. A horizontal shift of $\pi$ units to the left and a vertical shift of 2 units down.

Explanation: The transformations in A result in the function $g(x) = 3\tan(2x)$. To find the x-intercepts, we solve $3\tan(2x)=0$, which simplifies to $\tan(2x)=0$. This occurs when $2x = k\pi$, or $x = \frac{k\pi}{2}$. For $k=0$, there is an x-intercept at $x=0$. Horizontal and vertical shifts (choices B, C, D) will move the intercept from the origin to a new location.

Question 8

1. $(-\infty, \infty)$ (correct answer)
2. $[-1, 1]$
3. $(-\infty, -1] \cup [1, \infty)$
4. $(-\infty, 0) \cup (0, \infty)$

Explanation: The graph of the tangent function extends infinitely upwards and downwards between its vertical asymptotes. Therefore, the function can take on any real number value, and its range is all real numbers, or $(-\infty, \infty)$.

Question 9

1. The functions have the same zeros but different periods and ranges. (correct answer)
2. The functions have the same period but different zeros and asymptotes.
3. The functions have the same range but different periods and zeros.
4. The functions have the same domain and the same zeros but different periods.

Explanation: The zeros of $\tan(x)$ occur where $\sin(x)=0$, which is at $x=k\pi$ for any integer $k$. So, they have the same zeros. The period of $\tan(x)$ is $\pi$, while the period of $\sin(x)$ is $2\pi$. The range of $\tan(x)$ is $(-\infty, \infty)$, while the range of $\sin(x)$ is $[-1, 1]$. Thus, their periods and ranges are different.

Question 10

1. $\frac{\pi}{3}$
2. $3\pi$ (correct answer)
3. $\frac{5\pi}{3}$
4. $5\pi$

Explanation: The period of the parent function $y=\tan(x)$ is $\pi$. For a function of the form $y = a\tan(bx)$, the period is given by $\frac{\pi}{|b|}$. For $g(x) = 5\tan(\frac{1}{3}x)$, we have $b=\frac{1}{3}$. Therefore, the period is $\frac{\pi}{1/3} = 3\pi$.

Question 11

1. $\frac{\pi}{2}$ (correct answer)
2. $2\pi$
3. $\pi$
4. $\frac{\pi}{4}$

Explanation: For a function of the form $y = a\tan(b(x-c)) + d$, the period is given by $\frac{\pi}{|b|}$. The function can be rewritten as $f(x) = -4\tan(2(x - \frac{\pi}{2}))$ so $b=2$. The period is $\frac{\pi}{2}$.

Question 12

1. A vertical stretch by a factor of 2 and a reflection across the x-axis. (correct answer)
2. A vertical compression by a factor of 2 and a reflection across the y-axis.
3. A horizontal stretch by a factor of 2 and a reflection across the x-axis.
4. A vertical stretch by a factor of 2 and a translation 2 units down.

Explanation: For a function $y = af(x)$, the parameter $a$ causes a vertical stretch by a factor of $|a|$. If $a$ is negative, it also causes a reflection across the x-axis. In this case, $a=-2$, so there is a vertical stretch by a factor of 2 and a reflection across the x-axis.

Question 13

1. The function is always increasing on the entire interval. (correct answer)
2. The function is always decreasing on the entire interval.
3. The function increases on $(\frac{\pi}{2}, \pi)$ and then decreases on $(\pi, \frac{3\pi}{2})$.
4. The function decreases on $(\frac{\pi}{2}, \pi)$ and then increases on $(\pi, \frac{3\pi}{2})$.

Explanation: The interval $(\frac{\pi}{2}, \frac{3\pi}{2})$ is one full period of the tangent function, between two consecutive vertical asymptotes. Throughout any such interval, the tangent function is strictly increasing.

Question 14

1. $x=\dfrac{\pi}{2}+k\pi$
2. $x=\dfrac{\pi}{4}+k\pi$
3. $x=k\pi$ (correct answer)
4. $x=\dfrac{k\pi}{2}$

Explanation: This question tests AP Precalculus understanding of the tangent function's properties, specifically locating its zeros based on the quotient definition. Since tan x = sin x/cos x, the function equals zero when the numerator sin x = 0 and the denominator cos x ≠ 0, which occurs at integer multiples of π. In this question, students must identify where sin x = 0 while ensuring cos x ≠ 0 to avoid undefined points. Choice C is correct because x = kπ represents all integer multiples of π (0, ±π, ±2π, etc.), where sine equals zero and cosine equals ±1. Choice A is incorrect because x = π/2 + kπ represents the asymptotes where cos x = 0, not the zeros of tangent. To help students: Graph y = sin x, y = cos x, and y = tan x together to visualize where tangent crosses the x-axis. Emphasize that zeros occur where the numerator is zero but the denominator isn't, distinguishing zeros from undefined points.

Question 15

1. $x=k\pi$
2. $x=\dfrac{\pi}{2}+k\pi$ (correct answer)
3. $x=2k\pi$
4. $x=\dfrac{\pi}{2}+2k\pi$

Explanation: This question tests AP Precalculus understanding of the tangent function's properties, specifically identifying where vertical asymptotes occur. The tangent function is defined as the ratio of sine to cosine, and vertical asymptotes occur where the denominator (cosine) equals zero. In this question, students must determine when cos(x) = 0, which happens at odd multiples of π/2. Choice B is correct because x = π/2 + kπ represents all odd multiples of π/2 where k is any integer, precisely where cosine equals zero. Choice A is incorrect because it represents multiples of π where cosine alternates between 1 and -1, not zero. To help students: Emphasize that vertical asymptotes occur when denominators equal zero. Practice identifying zeros of cosine by visualizing the unit circle or cosine graph.

Question 16

1. $f(-x) = -f(x)$ (correct answer)
2. $f(-x) = f(x)$
3. $f(x) = f(x+\pi)$
4. $f(x) = \frac{1}{f(x)}$

Explanation: The definition of an odd function is that $f(-x) = -f(x)$ for all $x$ in its domain. This corresponds to symmetry about the origin, which the tangent function possesses. Choice B defines an even function. Choice C is the definition of a periodic function with period $\pi$, which is true for tangent but is not the definition of an odd function.

Question 17

1. At $x = k\pi$ for any integer $k$. (correct answer)
2. At $x = \frac{\pi}{2} + k\pi$ for any integer $k$.
3. At $x = \frac{\pi}{4} + \frac{k\pi}{2}$ for any integer $k$.
4. At $x = 2k\pi$ for any integer $k$.

Explanation: The zeros of $\tan(x) = \frac{\sin(x)}{\cos(x)}$ occur when the numerator, $\sin(x)$, is equal to 0, provided the denominator is not also 0. The function $\sin(x)$ is zero at all integer multiples of $\pi$. At these values, $\cos(x)$ is either 1 or -1, so the denominator is not zero.

Question 18

1. $x = 4\pi$
2. $x = 6\pi$ (correct answer)
3. $x = 3\pi$
4. $x = 10\pi$

Explanation: The period of $g(x) = \tan(\frac{1}{4}x)$ is $\frac{\pi}{|b|} = \frac{\pi}{1/4} = 4\pi$. The vertical asymptotes of a tangent function are separated by a distance equal to its period. If one asymptote is at $x=2\pi$, the next one for increasing $x$ will be at $x = 2\pi + \text{period} = 2\pi + 4\pi = 6\pi$.

Question 19

1. $\theta=\arctan\!\left(\dfrac{2}{3}\right)$ (correct answer)
2. $\theta=\arcsin\!\left(\dfrac{2}{3}\right)$
3. $\theta=\arccos\!\left(\dfrac{2}{3}\right)$
4. $\theta=\arctan\!\left(\dfrac{3}{2}\right)$

Explanation: This question tests AP Precalculus understanding of the tangent function's properties, specifically using inverse tangent to find angles from known ratios. The tangent function is defined as the ratio of sine to cosine, and its inverse function arctan returns the angle whose tangent equals a given value. In this question, the architectural context with a 20m tower viewed from 30m away creates tan θ = 20/30 = 2/3. Choice A is correct because θ = arctan(2/3) properly uses the inverse tangent function to find the angle whose tangent equals 2/3. Choice D is incorrect because it inverts the fraction to 3/2, which would represent the angle if viewing from 20m away at a 30m tower. To help students: Emphasize that arctan 'undoes' the tangent function to recover angles. Practice setting up the opposite/adjacent ratio correctly before applying inverse tangent.

Question 20

1. $x=k\pi$
2. $x=\dfrac{k\pi}{2}$
3. $x=\dfrac{\pi}{2}+k\pi$ (correct answer)
4. $x=2k\pi$

Explanation: This question tests AP Precalculus understanding of the tangent function's properties, specifically identifying where vertical asymptotes occur based on the function's definition. The tangent function is defined as sin θ/cos θ, which means it becomes undefined wherever cos θ = 0, creating vertical asymptotes at these points. In this question, students must identify all locations where cosine equals zero, which occurs at odd multiples of π/2. Choice C is correct because x = π/2 + kπ represents all odd multiples of π/2 (like π/2, 3π/2, 5π/2, etc.), which are precisely where cosine equals zero. Choice A is incorrect because x = kπ includes points like 0 and π where cosine equals 1 or -1, not zero. To help students: Draw the cosine graph and mark all zeros to visualize asymptote locations. Practice converting between different representations of periodic points, emphasizing that π/2 + kπ captures all odd multiples of π/2.