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Difficulty: Very hardGraphs of Trigonometric Functions

The graph of the function f(x)=acos⁡(b(x−c))+df(x) = a \cos(b(x - c)) + d is shown below for constants a>0a > 0, b>0b > 0, c∈[0,π]c \in [0, \pi], and dd. The graph has a local maximum at (π3,5)(\frac{\pi}{3}, 5) and the nearest local minimum to its right is at (5π6,−1)(\frac{5\pi}{6}, -1). What is the y-intercept of the graph of f(x)f(x)?

  1. 12\frac{1}{2}Answer
  2. B
    72\frac{7}{2}
  3. C
    −52-\frac{5}{2}
  4. D
    22
  5. E
    −72-\frac{7}{2}

Answer

The y-intercept of the graph is 12\frac{1}{2}
The correct answer is 12\frac{1}{2}. The amplitude of the function is a=5−(−1)2=3a = \frac{5 - (-1)}{2} = 3, and the midline is d=5+(−1)2=2d = \frac{5 + (-1)}{2} = 2. The distance between the consecutive local maximum and local minimum represents half of a period: T2=5π6−π3=π2\frac{T}{2} = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}, meaning the period is T=πT = \pi, so b=2ππ=2b = \frac{2\pi}{\pi} = 2. A local maximum occurs when the argument of the cosine function is 00, so 2(π3−c)=0  ⟹  c=π32(\frac{\pi}{3} - c) = 0 \implies c = \frac{\pi}{3}. This gives the equation f(x)=3cos⁡(2(x−π3))+2f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2. Finding the y-intercept requires evaluating the function at x=0x = 0: f(0)=3cos⁡(2(0−π3))+2=3cos⁡(−2π3)+2=3(−12)+2=12f(0) = 3 \cos(2(0 - \frac{\pi}{3})) + 2 = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}.

Step-by-Step Solution

1
Find the amplitude aa and midline dd of the trigonometric function.
a=3a = 3 and d=2d = 2
The amplitude is half the distance between the maximum and minimum values: a=5−(−1)2=3a = \frac{5 - (-1)}{2} = 3. The midline is the average of these values: d=5+(−1)2=2d = \frac{5 + (-1)}{2} = 2.
2
Determine the period TT and the frequency coefficient bb.
T=πT = \pi and b=2b = 2
The horizontal distance between a consecutive maximum and minimum is half of the period: 5π6−π3=π2\frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}. Thus, the full period is T=πT = \pi. Since T=2πbT = \frac{2\pi}{b}, we find b=2b = 2.
3
Determine the horizontal phase shift cc.
c=π3c = \frac{\pi}{3}
A cosine function achieves its maximum when its argument is a multiple of 2π2\pi. Since the maximum is at x=π3x = \frac{\pi}{3}, we set 2(π3−c)=02(\frac{\pi}{3} - c) = 0, giving c=π3c = \frac{\pi}{3}.
4
Evaluate the function at x=0x = 0 to find the y-intercept.
f(0)=12f(0) = \frac{1}{2}
Substitute the parameters into the function: f(x)=3cos⁡(2(x−π3))+2f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2. Substituting x=0x = 0 gives f(0)=3cos⁡(−2π3)+2=3(−12)+2=12f(0) = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}.

Key Concept

Determining the equation of a transformed trigonometric function from key features (maximum and minimum points) and evaluating it.
Estimated Time:3m 0s
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Topics

Graphs of Trigonometric Functions

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