Tangent Graphs: Period, Asymptotes, Shifts & Midlines Quiz

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| Questions: 20 | Updated: Jan 22, 2026

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1) A tangent function is used to model tide current speed v(t) with mean 0 and period 6 hours, steep increases near odd multiples of 3 hours. Which function fits?

V(t) = A tan(πt/6)

V(t) = A tan(2πt/6)

V(t) = A tan(πt/3)

V(t) = A tan(2πt/3)

Given: period 6 ⇒ π/|b| = 6 ⇒ b = π/6. Goal: choose model.

Step 1: v(t) = A tan(πt/6) has asymptotes at t = 3 + 6k (odd multiples of 3).

So, the final answer is v(t) = A tan(πt/6).

Explanation

Given: period 6 ⇒ π/|b| = 6 ⇒ b = π/6. Goal: choose model.

Step 1: v(t) = A tan(πt/6) has asymptotes at t = 3 + 6k (odd multiples of 3).

So, the final answer is v(t) = A tan(πt/6).

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About This Quiz

Tangent Graphs: Period, Asymptotes, Shifts & Midlines Quiz - Quiz

Ready to read tangent graphs like a pro? In this quiz, you’ll spot the period, find vertical asymptotes, and locate the “S-shaped” inflection point that sits on the midline. You’ll match equations to graphs and see how changes to the formula stretch, flip, and shift the curve. By the end,... see moreyou’ll be quick at identifying what each part of the equation does to the graph!
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2) The function y = 3 tan(2x) has which fundamental period?

π/2

π/3

2π

Given: y = 3 tan(2x). Goal: Find period.

Step 1: For y = tan(bx), period = π/|b|.

Step 2: b = 2 ⇒ period = π/2.

So, the final answer is π/2.

Explanation

Given: y = 3 tan(2x). Goal: Find period.

Step 1: For y = tan(bx), period = π/|b|.

Step 2: b = 2 ⇒ period = π/2.

So, the final answer is π/2.

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3) Which transformation maps y = tan(x) to y = −2 tan(x − π/4)?

Vertical stretch by factor 2, reflection across x-axis, and right shift by π/4

Horizontal stretch by factor 2, reflection across y-axis, and left shift by π/4

Vertical stretch by factor 2, reflection across y-axis, and right shift by π/4

Horizontal compression by factor 2, reflection across x-axis, and left shift by π/4

Given: target y = −2 tan(x − π/4). Goal: describe transformations.

Step 1: “−” ⇒ reflect across x-axis; “2·” ⇒ vertical stretch by 2.

Step 2: “(x − π/4)” ⇒ shift right by π/4.

So, the final answer is vertical stretch 2, reflect across x-axis, shift right π/4.

Explanation

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4) The graph has asymptotes at x = −π/4 and x = 3π/4 and passes through (π/4, 2). Which function?

Y = 2 tan(2x)

Y = 2 tan(x − π/4)

Y = 2 tan(2x + π/2)

Y = 2 tan(2x − π/2)

Given: asymptotes at −π/4 and 3π/4, midpoint π/4 with output 2. Goal: find model.

Step 1: For y = a tan(b(x − c)), asymptotes c ± π/(2b); midpoint is x = c.

Step 2: Midpoint c = π/4; distance between asymptotes = π ⇒ π/b = π ⇒ b = 1.

Step 3: Value at midpoint equals vertical shift (here 2·tan(0)=0) but they gave point (π/4, 2) meaning a = 2 scaling around midline 0 fits with y-value at other x, So, the final answer is y = 2 tan(x − π/4).

Explanation

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5) In y = a tan(b(x − c)) + d, consecutive asymptotes at x = 1 and x = 3. What is b?

π/2

1/π

Given: asymptote spacing = 3 − 1 = 2. Goal: find b.

Step 1: For tangent, spacing between consecutive asymptotes is π/|b|.

Step 2: π/|b| = 2 ⇒ b = π/2.

So, the final answer is π/2.

Explanation

Given: asymptote spacing = 3 − 1 = 2. Goal: find b.

Step 1: For tangent, spacing between consecutive asymptotes is π/|b|.

Step 2: π/|b| = 2 ⇒ b = π/2.

So, the final answer is π/2.

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6) The midline of y = tan(x) + 4 is at:

Y = 0

Y = 4

Y = −4

X = 4

Given: y = tan(x) + 4. Goal: midline.

Step 1: Vertical shift d = 4 sets midline y = 4.

So, the final answer is y = 4.

Explanation

Given: y = tan(x) + 4. Goal: midline.

Step 1: Vertical shift d = 4 sets midline y = 4.

So, the final answer is y = 4.

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7) A tangent graph shows a point of inflection at (2, −3), with adjacent vertical asymptotes at x = 1 and x = 3. Which equation matches?

Y = tan(π(x − 2)) − 3

Y = 2 tan(π(x − 2)) − 3

Y = tan(π/2 (x − 2)) − 3

Y = tan(π/2 (x − 2)) + 3

Given: center (c, d) = (2, −3), spacing 3 − 1 = 2. Goal: model.

Step 1: π/|b| = 2 ⇒ b = π/2.

Step 2: Use y = tan((π/2)(x − 2)) − 3.

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

Explanation

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8) Which equation has period π/3 and midline y = −2?

Y = tan(3x) − 2

Y = tan(x/3) − 2

Y = −2 tan(3x)

Y = 3 tan(x) − 2

Given: period π/3, midline −2. Goal: pick model.

Step 1: π/|b| = π/3 ⇒ b = 3.

Step 2: Add −2 for midline shift: y = tan(3x) − 2.

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

Explanation

Given: period π/3, midline −2. Goal: pick model.

Step 1: π/|b| = π/3 ⇒ b = 3.

Step 2: Add −2 for midline shift: y = tan(3x) − 2.

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

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9) A tangent model has vertical asymptotes at x = −π/6 and x = π/6, and crosses the midline at (0, −1). Which is the correct model?

Y = tan(3x) + 1

Y = tan(x/3) − 1

Y = 3 tan(x) − 1

Y = tan(3x) − 1

Given: spacing = π/3 ⇒ b = 3; midline −1 at x = 0. Goal: model.

Step 1: Use y = tan(3x) + d; midline crossing at x = 0 gives y = d = −1.

So, the final answer is y = tan(3x) − 1.

Explanation

Given: spacing = π/3 ⇒ b = 3; midline −1 at x = 0. Goal: model.

Step 1: Use y = tan(3x) + d; midline crossing at x = 0 gives y = d = −1.

So, the final answer is y = tan(3x) − 1.

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10) Consider y = −4 tan(0.5(x + π)) + 2. Which statement is true?

Period is 2π; midline y = 2; reflected across x-axis

Period is π; midline y = −2; reflected across y-axis

Period is π; midline y = 2; reflected across x-axis

Period is 2π; midline y = −2; no reflection

Given: b = 0.5, a = −4, d = 2. Goal: features.

Step 1: Period = π/|b| = π/0.5 = 2π; midline y = d = 2.

Step 2: Negative a ⇒ reflection across x-axis.

So, the final answer is period 2π, midline y = 2, reflection across x-axis.

Explanation

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11) A tangent function models temperature deviation T(t) from average over time t (hours). It peaks upward steeply at t = 0, crosses its midline at t = 1 with T = 0, and has vertical asymptotes at t = −0.5 and t = 1.5. Which equation fits best?

T(t) = tan(π(t − 1))

T(t) = tan(π/2 (t − 0.5))

T(t) = tan(π(t − 0.5))

T(t) = tan(π/2 (t − 1))

Given: asymptotes at −0.5 and 1.5, midpoint 0.5. Goal: model.

Step 1: For y = tan(b(t − c)), asymptotes at t = c ± π/(2b).

Step 2: Choose c = 0.5 and π/b = 2 ⇒ b = π/2.

So, the final answer is T(t) = tan(π/2 (t − 0.5)).

Explanation

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12) The graph of y = a tan(bx) has vertical asymptotes at x = ±π/8. What is b?

1/4

Given: asymptotes at ±π/8 ⇒ spacing π/4. Goal: b.

Step 1: π/|b| = π/4 ⇒ b = 4.

So, the final answer is 4.

Explanation

Given: asymptotes at ±π/8 ⇒ spacing π/4. Goal: b.

Step 1: π/|b| = π/4 ⇒ b = 4.

So, the final answer is 4.

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13) A tangent function has period π/4 and passes through (0, 0). Which is possible?

Y = 4 tan(4x)

Y = tan(x/4)

Y = tan(4x)

Y = tan(πx/4)

Given: period π/4. Goal: pick model.

Step 1: π/|b| = π/4 ⇒ b = 4.

Step 2: y = tan(4x) passes through (0,0).

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

Explanation

Given: period π/4. Goal: pick model.

Step 1: π/|b| = π/4 ⇒ b = 4.

Step 2: y = tan(4x) passes through (0,0).

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

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14) A graph looks like a standard tangent wave, shifted upward by 3 units, with midline y = 3, and period π/2. Which is the correct equation?

Y = tan(2x) + 3

Y = tan(x/2) + 3

Y = 2 tan(x) + 3

Y = tan(2x) − 3

Given: period π/2 ⇒ b = 2; vertical shift +3. Goal: model.

Step 1: Use y = tan(2x) + 3.

So, the final answer is y = tan(2x) + 3.

Explanation

Given: period π/2 ⇒ b = 2; vertical shift +3. Goal: model.

Step 1: Use y = tan(2x) + 3.

So, the final answer is y = tan(2x) + 3.

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15) Given y = 5 tan(3(x − π/6)) − 1. What are asymptotes in 0 < x < π?

X = π/6, π/2, 5π/6

X = π/6 ± π/6, π/2 ± π/6

X = π/6 + kπ/3 for integers k with result in (0, π)

X = kπ (0, π)

Given: c = π/6, b = 3. Goal: asymptotes.

Step 1: Asymptotes at x = c ± π/(2b) + kπ/b = π/6 ± π/6 + kπ/3.

Step 2: In (0, π): π/6, π/2, 5π/6.

So, the final answer is x = π/6, π/2, 5π/6.

Explanation

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16) Which statement about y = tan(bx) is true?

Increasing b stretches the graph horizontally.

Increasing b compresses the graph horizontally and decreases the period.

Increasing b increases the vertical stretch.

Increasing b shifts the midline.

Step 1: For y = tan(bx), the period of tangent is given by π ⁄ b.

Step 2: When b increases, π ⁄ b becomes smaller, so the graph completes one cycle more quickly.

Step 3: That means the graph is horizontally compressed (the period decreases).

So, the final answer is “Increasing b compresses the graph horizontally and decreases the period.”

Explanation

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17) The graph of y = −tan(x − π/2) is equivalent to which transformation of y = tan(x)?

Reflect across y-axis and shift right by π/2

Reflect across x-axis and shift right by π/2

Shift left by π/2 and reflect across y-axis

Shift right by π/2 only

Given: y = −tan(∙) and (x − π/2). Goal: transformation.

Step 1: (x − π/2) ⇒ shift right π/2; “−” ⇒ reflect across x-axis.

So, the final answer is reflect across x-axis and shift right by π/2.

Explanation

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18) For y = a tan(b(x − c)) + d, which parameter primarily determines the horizontal distance between consecutive vertical asymptotes?

Given: spacing = π/|b|. Goal: parameter.

Step 1: Only b affects period and spacing.

So, the final answer is b.

Explanation

Given: spacing = π/|b|. Goal: parameter.

Step 1: Only b affects period and spacing.

So, the final answer is b.

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19) A tangent graph has midline y = −2, inflection at (1, −2), asymptotes at x = 0 and x = 2. Which equation?

Y = tan((x − 1)/2) − 2

Y = tan(π(x − 1)) − 2

Y = −2 tan(π/2 (x − 1)) − 2

Y = tan(π/2 (x − 1)) − 2

Given: center (1, −2), spacing 2 ⇒ π/|b| = 2. Goal: model.

Step 1: b = π/2; c = 1; d = −2.

So, the final answer is y = tan(π/2 (x − 1)) − 2.

Explanation

Given: center (1, −2), spacing 2 ⇒ π/|b| = 2. Goal: model.

Step 1: b = π/2; c = 1; d = −2.

So, the final answer is y = tan(π/2 (x − 1)) − 2.

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20) A function has period π, vertical asymptotes at x = 3π/4 + kπ, and crosses the midline at x = π + kπ. Which could it be?

Y = tan(x − π/4)

Y = tan(x − 3π/4)

Y = tan(2x − 3π/2)

Y = tan(x + π/4)

Given: period π. Goal: match asymptotes and crossings.

Step 1: For y = tan(x − π/4), asymptotes occur where x − π/4 = ±π/2 + kπ ⇒ x = 3π/4 + kπ (matches).

Step 2: Midline crossings occur at x = π/4 + kπ (phase-shifted zeros); among given options, only A matches the asymptote condition and period.

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

Explanation

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