Question

Difficulty: MediumGraphs of Trigonometric Functions

Match each transformed trigonometric function on the left with its set of defining graphical properties on the right.

  • f(x)=4sin(3x)+2f(x) = 4\sin(3x) + 2Amplitude of 44, period of 2π3\frac{2\pi}{3}, and a midline at y=2y = 2
  • g(x)=2cos(2xπ)g(x) = -2\cos\left(2x - \pi\right)Amplitude of 22, period of π\pi, and a phase shift of π2\frac{\pi}{2} units to the right
  • h(x)=3tan(12x)1h(x) = 3\tan\left(\frac{1}{2}x\right) - 1Period of 2π2\pi, vertical shift of 11 unit down, with vertical asymptotes at x=π+2kπx = \pi + 2k\pi for any integer kk
  • k(x)=cos(x+π3)4k(x) = \cos\left(x + \frac{\pi}{3}\right) - 4Midline at y=4y = -4, amplitude of 11, and a phase shift of π3\frac{\pi}{3} units to the left

Answer

The correct pairings are: f(x)=4sin(3x)+2f(x) = 4\sin(3x) + 2 matches with 'Amplitude of 4, period of 2π/3, and a midline at y = 2'; g(x)=2cos(2xπ)g(x) = -2\cos(2x - π) matches with 'Amplitude of 2, period of π, and a phase shift of π/2 units to the right'; h(x)=3tan(x/2)1h(x) = 3\tan(x/2) - 1 matches with 'Period of 2π, vertical shift of 1 unit down, with vertical asymptotes at x = π + 2kπ'; and k(x)=cos(x+π/3)4k(x) = \cos(x + π/3) - 4 matches with 'Midline at y = -4, amplitude of 1, and a phase shift of π/3 units to the left'.
Each trigonometric function is matched according to its standard parameter transformation rules: y=Asin(B(xC))+Dy = A\sin(B(x - C)) + D or y=Acos(B(xC))+Dy = A\cos(B(x - C)) + D, where A|A| is amplitude, period is 2πB\frac{2\pi}{|B|} for sine/cosine and πB\frac{\pi}{|B|} for tangent, CC is horizontal phase shift, and y=Dy = D is the midline.

Step-by-Step Solution

1
Analyze f(x)=4sin(3x)+2f(x) = 4\sin(3x) + 2 using the standard form y=Asin(B(xC))+Dy = A\sin(B(x - C)) + D
Amplitude =A=4= |A| = 4, period =2πB=2π3= \frac{2\pi}{B} = \frac{2\pi}{3}, midline =D    y=2= D \implies y = 2.
Direct extraction of parameters for sine graphs.
2
Factor out B=2B = 2 from g(x)=2cos(2xπ)g(x) = -2\cos(2x - \pi)
g(x)=2cos(2(xπ2))g(x) = -2\cos\left(2\left(x - \frac{\pi}{2}\right)\right), so amplitude =2=2= |-2| = 2, period =2π2=π= \frac{2\pi}{2} = \pi, phase shift =π2= \frac{\pi}{2} to the right.
Factoring BB is necessary to correctly identify the horizontal phase shift.
3
Analyze tangent function parameters for h(x)=3tan(12x)1h(x) = 3\tan\left(\frac{1}{2}x\right) - 1
Period =πB=π1/2=2π= \frac{\pi}{B} = \frac{\pi}{1/2} = 2\pi, shifted down 11 unit (y=1y = -1). Asymptotes occur when 12x=π2+kπ    x=π+2kπ\frac{1}{2}x = \frac{\pi}{2} + k\pi \implies x = \pi + 2k\pi.
Tangent period uses πB\frac{\pi}{B} instead of 2πB\frac{2\pi}{B}, and asymptotes occur where tangent arguments equal odd multiples of π2\frac{\pi}{2}.
4
Analyze transformation parameters for k(x)=cos(x+π3)4k(x) = \cos\left(x + \frac{\pi}{3}\right) - 4
Amplitude =1= 1, midline =y=4= y = -4, phase shift =π3= \frac{\pi}{3} to the left.
Addition inside the function argument (x+C)(x + C) corresponds to a horizontal shift to the left.

Key Concept

Identifying amplitude, period, midline, phase shift, and asymptotes from transformed trigonometric equations
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