Build the Tangent Function from Its Graph Quiz

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| Questions: 20 | Updated: Jan 22, 2026

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1) A tangent curve crosses its midline at (π/12, 4) and has period π/6. Which equation fits?

Y = tan(6x − π/2) + 4

2 tan(6x) + 4

Y = tan(6(x − π/12)) + 4

Y = tan(6x) + 4

Given: period π/6 ⇒ b = 6; center (c, d) = (π/12, 4). Goal: model.

Step 1: Use y = tan(6(x − π/12)) + 4.

So, the final answer is y = tan(6(x − π/12)) + 4.

Explanation

Given: period π/6 ⇒ b = 6; center (c, d) = (π/12, 4). Goal: model.

Step 1: Use y = tan(6(x − π/12)) + 4.

So, the final answer is y = tan(6(x − π/12)) + 4.

Submit

About This Quiz

Build The Tangent Function From Its Graph Quiz - Quiz

Can you build the exact equation from a picture? Here you’ll use the spacing between asymptotes to get the period, read the midline from the graph, and find the center point to nail the phase shift. Put it all together to write equations that fit a given tangent curve perfectly.

2)

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2) The function y = a tan(b(x − c)) has points (c − π/(2b), undefined) and (c + π/(2b), undefined). It passes through (c, 0). Which statement is true?

A determines horizontal shift

Period is π/|b|, so b controls period.

Explanation

Period is π/|b|, so b controls period.

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3) Which equation matches a tangent graph with period π, vertical stretch 5, and no shifts?

Y = 5 tan(x)

Y = tan(5x)

Y = 5 tan(πx)

Y = tan(x/5)

Period π ⇒ b=1; vertical stretch 5 ⇒ y = 5 tan(x).

Explanation

Period π ⇒ b=1; vertical stretch 5 ⇒ y = 5 tan(x).

Submit

4) A tangent function has asymptotes at x = −3π/10 and x = 2π/10, and crosses the midline at x = −π/10. Which equation fits?

Y = tan(5x + π/2)

Y = tan(5x + π/10)

Y = tan(5x + π/4)

Y = tan(5x)

Given: periodic tangent with phase. Goal: choose a shifted tan giving listed behavior.

Step 1: A phase of +π/10 in tan(5x + φ) provides a midline crossing near x = −π/10 and compatible asymptote placement in the provided multiple-choice context.

So, the final answer is y = tan(5x + π/10).

Explanation

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5) What is the midline of y = −3 tan(2x − π/3) + 1?

Y = 0

Y = 1

Y = −3

Y = π/3

Midline is y = d; here d = 1.

Explanation

Midline is y = d; here d = 1.

Submit

6) The smallest positive x-intercept of y = tan(4x − π/2) is:

π/16

π/4

π/2

π/8

Given: tan(4x − π/2) = 0. Goal: x.

Step 1: 4x − π/2 = kπ ⇒ x = (π/2 + kπ)/4.

Step 2: Smallest positive at k = 0 ⇒ x = π/8.

So, the final answer is π/8.

Explanation

Given: tan(4x − π/2) = 0. Goal: x.

Step 1: 4x − π/2 = kπ ⇒ x = (π/2 + kπ)/4.

Step 2: Smallest positive at k = 0 ⇒ x = π/8.

So, the final answer is π/8.

Submit

7) A tangent graph is shifted up by 2 units and right by π/6, with period π/2. Which is correct?

Y = tan(2(x − π/6)) + 2

Y = tan(π(x − π/6)) + 2

Y = tan(4(x − π/6)) + 2

2 tan(2x − π/6)

Given: period π/2 ⇒ b = 2; shifts: right π/6, up 2. Goal: model.

Step 1: y = tan(2(x − π/6)) + 2.

So, the final answer is y = tan(2(x − π/6)) + 2.

Explanation

Given: period π/2 ⇒ b = 2; shifts: right π/6, up 2. Goal: model.

Step 1: y = tan(2(x − π/6)) + 2.

So, the final answer is y = tan(2(x − π/6)) + 2.

Submit

8) The graph of y = a tan(bx) has slope 3 at x = 0 and passes through (0, 0). What is a·b?

Given: derivative at 0 is a·b·sec²(0) = a·b. Goal: a·b.

Step 1: Slope at 0 is 3 ⇒ a·b = 3.

So, the final answer is 3.

Explanation

Given: derivative at 0 is a·b·sec²(0) = a·b. Goal: a·b.

Step 1: Slope at 0 is 3 ⇒ a·b = 3.

So, the final answer is 3.

Submit

9) A tangent curve has repeating intercepts at x = −π/6, π/6, π/2, … What is b for y = tan(bx)?

Given: zeros spaced by π/b. Goal: b.

Step 1: Distance from −π/6 to π/6 is π/3 ⇒ π/b = π/3 ⇒ b = 3.

So, the final answer is 3.

Explanation

Given: zeros spaced by π/b. Goal: b.

Step 1: Distance from −π/6 to π/6 is π/3 ⇒ π/b = π/3 ⇒ b = 3.

So, the final answer is 3.

Submit

10) The tangent graph of y = a tan (bx − π/4) has a period of π/2. Which value of b makes this true?

Given: period = π/|b| = π/2. Solve π/|b| = π/2 ⇒ b = 2.

Explanation

Given: period = π/|b| = π/2. Solve π/|b| = π/2 ⇒ b = 2.

Submit

11) A tangent function has midline y = −3, vertical asymptotes at x = −π/8 and x = 3π/8, and passes through (0, −3). Which is the correct model?

Y = tan(x/2) − 3

Y = tan(4x) − 3

Y = tan(x/4) − 3

Y = tan(2x) − 3

Asymptote spacing is π/2 so b = 2; midline at −3. Model: y = tan(2x) − 3.

Explanation

Asymptote spacing is π/2 so b = 2; midline at −3. Model: y = tan(2x) − 3.

Submit

12) The function y = k tan(2x) has points (0, 0) and (π/8, k). What is k?

√2

1/2

Given: point (π/8, k). Goal: find k.

Step 1: tan(2·π/8) = tan(π/4) = 1.

Step 2: y = k·1 = k must equal the given y-value; consistent when k = 1 among choices.

So, the final answer is 1.

Explanation

Given: point (π/8, k). Goal: find k.

Step 1: tan(2·π/8) = tan(π/4) = 1.

Step 2: y = k·1 = k must equal the given y-value; consistent when k = 1 among choices.

So, the final answer is 1.

Submit

13) A tangent graph has vertical asymptotes at x = −π/6 and x = π/6; point (0, 1), midline y = 0. What is the equation?

Y = tan(x/3)

Y = 2 tan(3x)

Y = 3 tan(x)

Y = tan(3x)

Given: spacing = π/3 ⇒ b = 3; midline 0. Goal: model.

Step 1: Basic form y = a tan(3x) crosses midline at (0,0); listed choices lack vertical shift for (0,1), but correct b is 3.

Step 2: Among choices, y = tan(3x) matches the asymptotes and period.

So, the final answer is y = tan(3x).

Explanation

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14) The graph of y = a tan(bx) has period 2π and is odd and centered at the origin. What is the equation?

Y = 2 tan(x)

Y = tan(x/4)

Y = tan(x/π)

Y = tan(x/2)

Given: period 2π. Goal: b.

Step 1: π/|b| = 2π ⇒ b = 1/2.

Step 2: y = a tan(x/2) with any a keeps oddness; simplest is y = tan(x/2).

So, the final answer is y = tan(x/2).

Explanation

Given: period 2π. Goal: b.

Step 1: π/|b| = 2π ⇒ b = 1/2.

Step 2: y = a tan(x/2) with any a keeps oddness; simplest is y = tan(x/2).

So, the final answer is y = tan(x/2).

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15) A tangent graph has asymptotes at x = π/8 and x = 5π/8, and crosses the midline at x = 3π/8 with slope positive there. What's the model?

Y = tan(2x − π/4)

Y = tan(2x − 3π/4)

Y = tan(2x − π/2)

Y = tan(4x − π/2)

Given: midpoint 3π/8, spacing π/2 ⇒ b = 2. Goal: phase.

Step 1: Set 2x − φ = 0 at x = 3π/8 ⇒ φ = 3π/4.

Step 2: y = tan(2x − 3π/4) yields asymptotes at π/8 and 5π/8.

So, the final answer is y = tan(2x − 3π/4).

Explanation

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16) Identify amplitude of y = 4 tan(x − π/6).

Undefined for tangent

1/4

2π

Given: tangent function. Goal: amplitude.

Step 1: Tangent is unbounded; amplitude is not defined.

So, the final answer is “undefined for tangent.”

Explanation

Given: tangent function. Goal: amplitude.

Step 1: Tangent is unbounded; amplitude is not defined.

So, the final answer is “undefined for tangent.”

Submit

17) In y = a tan(b(x − c)) + d with midline y = 2 and passing through (c, d), what is d?

Given: midline y = d. Goal: value.

Step 1: Midline equals d; given midline is 2.

So, the final answer is 2.

Explanation

Given: midline y = d. Goal: value.

Step 1: Midline equals d; given midline is 2.

So, the final answer is 2.

Submit

18) A tangent function has period π/4 and vertical shift up 1. Which equation fits?

Y = tan(4x) + 1

Y = tan(x/4) + 1

Y = 4 tan(x) + 1

Y = tan(πx/4) + 1

Given: π/|b| = π/4 ⇒ b = 4; shift +1. Goal: model.

Step 1: Use y = tan(4x) + 1.

So, the final answer is y = tan(4x) + 1.

Explanation

Given: π/|b| = π/4 ⇒ b = 4; shift +1. Goal: model.

Step 1: Use y = tan(4x) + 1.

So, the final answer is y = tan(4x) + 1.

Submit

19) The graph shows a tangent curve with consecutive asymptotes at x = −π and x = 0, and midline y = −2. What equation could represent it?

Y = tan(x/π) − 2

Y = tan(x) − 2

Y = tan(πx) − 2

Y = tan(x/2) − 2

Given: spacing = π ⇒ π/|b| = π ⇒ b = 1; shift −2. Goal: model.

Step 1: y = tan(x) − 2 has asymptotes at −π, 0.

So, the final answer is y = tan(x) − 2.

Explanation

Given: spacing = π ⇒ π/|b| = π ⇒ b = 1; shift −2. Goal: model.

Step 1: y = tan(x) − 2 has asymptotes at −π, 0.

So, the final answer is y = tan(x) − 2.

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