Examkin

Question

Difficulty: HardGraphs of Trigonometric Functions

Match each transformed trigonometric function listed on the left with the correct description of its key graphical features listed on the right.

  • f(x)=−3cos⁡(2x−π2)+1f(x) = -3\cos\left(2x - \frac{\pi}{2}\right) + 1Has a period of π\pi, a midline at y=1y = 1, and a maximum value of 44 at x=3π4x = \frac{3\pi}{4}.
  • g(x)=2sin⁡(12x+π)−1g(x) = 2\sin\left(\frac{1}{2}x + \pi\right) - 1Has a period of 4π4\pi, a phase shift of 2π2\pi units to the left, and an absolute minimum value of −3-3 at x=πx = \pi.
  • h(x)=−tan⁡(3x+3π4)+2h(x) = -\tan\left(3x + \frac{3\pi}{4}\right) + 2Has a period of π3\frac{\pi}{3}, a y-intercept at (0,3)(0, 3), and consecutive vertical asymptotes spaced π3\frac{\pi}{3} units apart.
  • k(x)=4cos⁡(πx−π2)−3k(x) = 4\cos\left(\pi x - \frac{\pi}{2}\right) - 3Has a period of 22, a phase shift of 12\frac{1}{2} unit to the right, and a midline at y=−3y = -3.

Answer

The trigonometric functions correctly match their graphical features as follows: f(x)f(x) matches the description with period π\pi and midline y=1y = 1; g(x)g(x) matches the description with period 4π4\pi and phase shift 2π2\pi left; h(x)h(x) matches the description with period π/3\pi/3 and y-intercept (0,3)(0, 3); k(x)k(x) matches the description with period 22 and midline y=−3y = -3.
Each trigonometric equation is mapped to its unique set of graphical properties by evaluating its period, midline, phase shift, and specific points like y-intercepts or extrema using standard trigonometric transformation formulas.

Step-by-Step Solution

1
Analyze f(x)=−3cos⁡(2x−π2)+1f(x) = -3\cos\left(2x - \frac{\pi}{2}\right) + 1.
Factor out the coefficient of xx: f(x)=−3cos⁡(2(x−π4))+1f(x) = -3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1. The period is 2πB=2π2=π\frac{2\pi}{B} = \frac{2\pi}{2} = \pi. The midline is y=D=1y = D = 1. The maximum value occurs when the cosine term equals −1-1 (due to the −3-3 coefficient): −3(−1)+1=4-3(-1) + 1 = 4, which happens at 2x−π2=π  ⟹  x=3π42x - \frac{\pi}{2} = \pi \implies x = \frac{3\pi}{4}.
Identify period, midline, phase shift, and extrema from standard form y=Acos⁡(B(x−C))+Dy = A\cos(B(x-C)) + D.
2
Analyze g(x)=2sin⁡(12x+π)−1g(x) = 2\sin\left(\frac{1}{2}x + \pi\right) - 1.
Rewrite as g(x)=2sin⁡(12(x−(−2π)))−1g(x) = 2\sin\left(\frac{1}{2}(x - (-2\pi))\right) - 1. The period is 2π1/2=4π\frac{2\pi}{1/2} = 4\pi, and the phase shift is 2π2\pi units to the left. The minimum value is 2(−1)−1=−32(-1) - 1 = -3, which occurs when 12x+π=3π2  ⟹  x=π\frac{1}{2}x + \pi = \frac{3\pi}{2} \implies x = \pi.
Determine horizontal shift, period, and minimum location.
3
Analyze h(x)=−tan⁡(3x+3π4)+2h(x) = -\tan\left(3x + \frac{3\pi}{4}\right) + 2.
The period for tangent is πB=π3\frac{\pi}{B} = \frac{\pi}{3}. Consecutive vertical asymptotes occur every period π3\frac{\pi}{3}. Evaluating at x=0x = 0 gives h(0)=−tan⁡(3π4)+2=−(−1)+2=3h(0) = -\tan\left(\frac{3\pi}{4}\right) + 2 = -(-1) + 2 = 3, giving a y-intercept of (0,3)(0, 3).
Apply tangent period formula π∣B∣\frac{\pi}{|B|} and evaluate y-intercept.
4
Analyze k(x)=4cos⁡(πx−π2)−3k(x) = 4\cos\left(\pi x - \frac{\pi}{2}\right) - 3.
Rewrite as k(x)=4cos⁡(π(x−12))−3k(x) = 4\cos\left(\pi\left(x - \frac{1}{2}\right)\right) - 3. The period is 2ππ=2\frac{2\pi}{\pi} = 2. The phase shift is 12\frac{1}{2} unit to the right, and the midline is y=−3y = -3.
Extract parameters from cosine function with π\pi in argument.

Key Concept

Graphical transformations of trigonometric functions (amplitude, period T=2π∣B∣T = \frac{2\pi}{|B|} or π∣B∣\frac{\pi}{|B|}, phase shift CC, and midline DD).
Estimated Time:2m 0s
Rate this question

Topics

Graphs of Trigonometric Functions

More questions from this topic