Build The Secant Function From Its Graph Quiz

1) A secant graph has period π, vertical asymptotes at x = −3π/4, −π/4, π/4, 3π/4, … and a minimum value of y = −2 at x = 0.

Y = −2 sec(2x)

Y = −2 sec(4x)

Y = 2 sec(2x)

Y = −2 sec(2x − π)

Step 1: Period π ⇒ 2π/|b|=π ⇒ b=2.

Step 2: Minimum y = −2 at x=0 ⇒ negative scale of 2.

So, y = −2 sec(2x).

Explanation

Step 1: Period π ⇒ 2π/|b|=π ⇒ b=2.

Step 2: Minimum y = −2 at x=0 ⇒ negative scale of 2.

So, y = −2 sec(2x).

2) The graph of y = sec(x) is shifted right by π/3 and reflected across the x-axis. What is the resulting function?

Y = sec(x + π/3)

Y = sec(x − π/3)

Y = −sec(x − π/3)

Y = −sec(x + π/3)

Step 1: Right shift ⇒ (x − π/3).

Step 2: Reflection across x-axis ⇒ negative sign outside.

Explanation

Step 1: Right shift ⇒ (x − π/3).

Step 2: Reflection across x-axis ⇒ negative sign outside.

3) A secant graph has vertical asymptotes at x = π/6 + kπ/3 and y-intercept (0, 4). What is the function?

Y = 4 sec(3x − π/3)

Y = 4 sec(3x − π/2)

Y = 4 sec(3x + π/2)

Y = 4 sec(3x + π/3)

Step 1: Asymptotes at π/6 + kπ/3 ⇒ 3x = π/2 + kπ ⇒ phase −π/3.

Step 2: Intercept 4 ⇒ scale 4.

Explanation

Step 1: Asymptotes at π/6 + kπ/3 ⇒ 3x = π/2 + kπ ⇒ phase −π/3.

Step 2: Intercept 4 ⇒ scale 4.

4) Which transformation turns y = sec(x) into y = sec(2x − π)?

Horizontal stretch by 2, right shift π

Horizontal compression by 2, right shift π/2

Horizontal compression by 2, right shift π

Horizontal stretch by 2, left shift π/2

Step 1: Factor 2 ⇒ horizontal compression by 2.

Step 2: 2x − π = 2(x − π/2) ⇒ shift right π/2.

Explanation

Step 1: Factor 2 ⇒ horizontal compression by 2.

Step 2: 2x − π = 2(x − π/2) ⇒ shift right π/2.

5) A graph has vertical asymptotes at x = −π/2, π/2, 3π/2, … and local minima y = 1 at x = 0, 2π, ….

Y = sec(x)

Y = sec(x − π/2)

Y = 2 sec(x)

Y = sec(x + π)

Step 1: Asymptotes at ±π/2 match sec(x).

Step 2: Vertex (minimum) 1 at x=0 fits sec(x).

Explanation

Step 1: Asymptotes at ±π/2 match sec(x).

Step 2: Vertex (minimum) 1 at x=0 fits sec(x).

6) The function y = −3 sec(x − π/4) has which of the following features?

Period 2π; asymptotes x = π/2 + kπ; minimum branch passes (π/4, −3)

Period 2π; asymptotes x = π/2 + kπ; maximum branch passes (π/4, −3)

Period 2π; asymptotes x = π/2 + kπ; maximum branch passes (π/4, 3)

Period π; asymptotes x = π/4 + kπ/2; maximum branch passes (π/4, −3)

Step 1: At x=π/4 the argument is 0 ⇒ sec(0)=1.

Step 2: y=−3 ⇒ on an inverted (maximum) branch.

Explanation

Step 1: At x=π/4 the argument is 0 ⇒ sec(0)=1.

Step 2: y=−3 ⇒ on an inverted (maximum) branch.

7) A secant function has a period of 4π and minimum y = 2 at x = π.

Y = 2 sec(x/2 − π/2)

Y = 2 sec(x/2 − π)

Y = 2 sec(x/2)

Y = −2 sec(x/2 − π)

Step 1: Period 4π ⇒ b=1/2.

Step 2: Minimum at x=π ⇒ set (x/2 − π/2)=0 ⇒ shift π/2.

Explanation

Step 1: Period 4π ⇒ b=1/2.

Step 2: Minimum at x=π ⇒ set (x/2 − π/2)=0 ⇒ shift π/2.

8) A reciprocal trig graph has asymptotes at x = kπ and U-branches centered at x = π/2 + 2πk.

Y = sec(2x)

Y = sec(x)

Y = csc(x)

Y = csc(2x)

Step 1: csc(x) has asymptotes where sin(x)=0 ⇒ x = kπ.

Step 2: Vertices at x = π/2 + 2πk match csc(x).

Explanation

Step 1: csc(x) has asymptotes where sin(x)=0 ⇒ x = kπ.

Step 2: Vertices at x = π/2 + 2πk match csc(x).

9) Which is the correct domain for y = sec(3x − π/2)?

All real x except x = π/6 + kπ/3

All real x except x = π/3 + 2kπ/3

All real x except x = π/2 + kπ

All real x except x = (π/2 + kπ)/3

Step 1: Denominator zero when cos(3x − π/2)=0 ⇒ 3x − π/2 = π/2 + kπ.

Step 2: Solve ⇒ x = (π/2 + kπ)/3.

Explanation

Step 1: Denominator zero when cos(3x − π/2)=0 ⇒ 3x − π/2 = π/2 + kπ.

Step 2: Solve ⇒ x = (π/2 + kπ)/3.

10) The function y = a sec(bx + c) has period π and asymptotes at x = −π/4, π/4.

A any, b = 4, c = 0

A any, b = 2, c = −π/2

A any, b = 4, c = −π

A any, b = 2, c = 0

Step 1: Distance between asymptotes is π/2 ⇒ b = 2.

Step 2: Centered at 0 ⇒ c = 0.

Explanation

Step 1: Distance between asymptotes is π/2 ⇒ b = 2.

Step 2: Centered at 0 ⇒ c = 0.

11) A secant curve has asymptotes at x = −π, 0, π, … and a minimum y = −1 at x = −π/2, 3π/2, ….

Y = −sec(x)

Y = sec(2x)

Y = sec(x − π/2)

Y = sec(x + π/2)

Step 1: Reflecting sec(x) across x-axis gives minima at y = −1.

Step 2: Asymptotes remain at kπ/2 offsets.

Explanation

Step 1: Reflecting sec(x) across x-axis gives minima at y = −1.

Step 2: Asymptotes remain at kπ/2 offsets.

12) Which statement about y = 5 sec(2x) is true?

Period 2π; range ≤ −5 or ≥ 5

Period π; range ≤ −5 or ≥ 5

Period π; range ≤ −1/5 or ≥ 1/5

Period 2π; range ≤ −1/5 or ≥ 1/5

Step 1: Period = 2π/2 = π.

Step 2: Vertical scale 5 ⇒ range ≤ −5 or ≥ 5.

Explanation

Step 1: Period = 2π/2 = π.

Step 2: Vertical scale 5 ⇒ range ≤ −5 or ≥ 5.

13) For y = −2 sec(3x + π/3), what is one vertical asymptote?

X = −π/6

X = −π/18

X = π/18

X = −π/9

cos(3x + π/3) = 0 ⇒ 3x + π/3 = π/2 ⇒ x = π/18.

Explanation

cos(3x + π/3) = 0 ⇒ 3x + π/3 = π/2 ⇒ x = π/18.

14) A secant function has minimum y = 4 at x = 0 and asymptotes at x = ±π/2.

Y = 4 sec(x)

Y = −4 sec(x)

Y = 4 sec(2x)

Y = 4 sec(x − π/2)

sec(x) has vertex 1 at x=0 ⇒ scaled to 4 gives minimum 4.

Explanation

sec(x) has vertex 1 at x=0 ⇒ scaled to 4 gives minimum 4.

15) Which graph corresponds to y = sec(x) shifted left by π/3 and stretched by 3?

Asymptotes x = −π/2 − π/3 + kπ; vertices x = −π/3 + 2kπ; y = ±3

Asymptotes x = π/2 − π/3 + kπ; vertices x = −π/3 + 2kπ; y = ±3

Asymptotes x = π/2 + π/3 + kπ; vertices x = π/3 + 2kπ; y = ±3

Asymptotes x = −π/2 + π/3 + kπ; vertices x = −π/3 + 2kπ; y = ±3

Step 1: Left shift by π/3: subtract π/3 from all key x-values.

Step 2: Vertical stretch 3 sets vertex y = ±3.

Explanation

Step 1: Left shift by π/3: subtract π/3 from all key x-values.

Step 2: Vertical stretch 3 sets vertex y = ±3.

16) A graph shows “cup” branches with minima y = 2 at x = π/4 + 2πk, and asymptotes x = −π/4 + πk.

Y = 2 sec(2x − π/2)

Y = 2 sec(x + π/4)

Y = −2 sec(x − π/4)

Y = 2 sec(x − π/4)

Step 1: Minimum at x = π/4 suggests argument zero ⇒ (x − π/4).

Step 2: Scale 2 ⇒ amplitude 2.

Explanation

Step 1: Minimum at x = π/4 suggests argument zero ⇒ (x − π/4).

Step 2: Scale 2 ⇒ amplitude 2.

17) For y = a sec(bx), which yields asymptotes every π/6 units?

A = 1, b = 6

A = any, b = 3

A = 2, b = 6

A = 1, b = 2

Step 1: Asymptote spacing = π/|b|.

Step 2: π/b = π/6 ⇒ b = 6.

Explanation

Step 1: Asymptote spacing = π/|b|.

Step 2: π/b = π/6 ⇒ b = 6.

18) A secant curve has period 2π/5 and smallest positive asymptote at x = π/10.

Y = sec(5x − π/2)

Y = sec(5x)

Y = sec(5x − π)

Y = sec(5x + π/2)

Step 1: Period 2π/b = 2π/5 ⇒ b=5.

Step 2: Asymptotes of sec(5x) at x = π/10 + kπ/5; smallest positive is π/10.

Explanation

Step 1: Period 2π/b = 2π/5 ⇒ b=5.

Step 2: Asymptotes of sec(5x) at x = π/10 + kπ/5; smallest positive is π/10.

19) Identify the cosine base of y = sec(2x − π/2).

Y = cos(2x − π/2)

Y = 1/cos(2x − π/2)

Y = cos(2x + π/2)

Y = cos(2x − π/2) with amplitude 1/2

Secant is the reciprocal of cosine ⇒ base cosine is cos(2x − π/2).

Explanation

Secant is the reciprocal of cosine ⇒ base cosine is cos(2x − π/2).

20) A plotted function has asymptotes at x = kπ/2 and a vertex at (0, −1).

Y = −sec(x/2)

Y = sec(2x)

Y = −sec(x)

Y = −sec(2x)

Standard vertex at (0,1) for sec(x); multiply by −1 ⇒ (0, −1).

Explanation

Standard vertex at (0,1) for sec(x); multiply by −1 ⇒ (0, −1).

Build the Secant Function from Its Graph Quiz

1) A secant graph has period π, vertical asymptotes at x = −3π/4, −π/4, π/4, 3π/4, … and a minimum value of y = −2 at x = 0.

2)

What first name or nickname would you like us to use?

2) The graph of y = sec(x) is shifted right by π/3 and reflected across the x-axis. What is the resulting function?

3) A secant graph has vertical asymptotes at x = π/6 + kπ/3 and y-intercept (0, 4). What is the function?

4) Which transformation turns y = sec(x) into y = sec(2x − π)?

5) A graph has vertical asymptotes at x = −π/2, π/2, 3π/2, … and local minima y = 1 at x = 0, 2π, ….

6) The function y = −3 sec(x − π/4) has which of the following features?

7) A secant function has a period of 4π and minimum y = 2 at x = π.

8) A reciprocal trig graph has asymptotes at x = kπ and U-branches centered at x = π/2 + 2πk.

9) Which is the correct domain for y = sec(3x − π/2)?

10) The function y = a sec(bx + c) has period π and asymptotes at x = −π/4, π/4.

11) A secant curve has asymptotes at x = −π, 0, π, … and a minimum y = −1 at x = −π/2, 3π/2, ….

12) Which statement about y = 5 sec(2x) is true?

13) For y = −2 sec(3x + π/3), what is one vertical asymptote?

14) A secant function has minimum y = 4 at x = 0 and asymptotes at x = ±π/2.

15) Which graph corresponds to y = sec(x) shifted left by π/3 and stretched by 3?

16) A graph shows “cup” branches with minima y = 2 at x = π/4 + 2πk, and asymptotes x = −π/4 + πk.

17) For y = a sec(bx), which yields asymptotes every π/6 units?

18) A secant curve has period 2π/5 and smallest positive asymptote at x = π/10.

19) Identify the cosine base of y = sec(2x − π/2).

20) A plotted function has asymptotes at x = kπ/2 and a vertex at (0, −1).