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Item 1. Formula Sheet-PC Math 12. Below is the formula sheet for this course, as found in the course intro. You will be provided this within all quizzes and tests for your use. You can click back to this location to make use of the information as needed throughout the test.
This item is a formula sheet/reference item. No calculation is required. Use the formula sheet as needed throughout the quiz. Answer: N/N
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Item 2. Determine the horizontal translation of the function \(y=3\sin\left(2x-\frac{\pi}{4}\right)\).
To determine the horizontal translation, factor the coefficient of \(x\) inside the sine function. \(y=3\sin\left(2x-\frac{\pi}{4}\right)\) \(y=3\sin\left(2\left(x-\frac{\pi}{8}\right)\right)\) The expression is now in the form \(y=a\sin(b(x-c))+d\). Here \(c=\frac{\pi}{8}\), so the horizontal translation is \(\frac{\pi}{8}\) to the right. Answer: D
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Item 3. The range of trigonometric function \(y=a\cos x+b\) is \(-2\leq y\leq 8\). Determine the value of \(b\).
In \(y=a\cos x+b\), the value \(b\) is the vertical translation or equilibrium line. The equilibrium line is the average of the maximum and minimum values. \(b=\frac{\text{max}+\text{min}}{2}\) The maximum value is \(8\), and the minimum value is \(-2\). \(b=\frac{8+(-2)}{2}\) \(b=\frac{6}{2}=3\) Therefore \(b=3\). Answer: C
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Item 4. Which of the following has a period of \(7\)?
For a sinusoidal function written as \(y=\sin\left(\frac{2\pi(x-c)}{\text{period}}\right)+d\), the denominator under \(2\pi(x-c)\) gives the period. Option A is \(y=\sin\left(\frac{2\pi(x-3)}{7}\right)+4\). This is in the form with period \(7\). Therefore the function in option A has a period of \(7\). Answer: A
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Item 5. Given \(y=\sin\left(\frac{1}{2}x\right)\), what is the period of this function?
For \(y=\sin(nx)\), the period is \(\frac{2\pi}{|n|}\). Here \(n=\frac{1}{2}\). \(\text{Period}=\frac{2\pi}{\frac{1}{2}}\) \(\text{Period}=2\pi\cdot 2=4\pi\) Therefore the period is \(4\pi\). Answer: B
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Item 6. A mass is supported by a spring so that it rests \(30\text{ cm}\) above a table top. The mass is pulled down to a height of \(10\text{ cm}\) above the table top and released at time \(t=0\). It takes \(0.8\) seconds for the mass to reach a maximum height of \(50\text{ cm}\) above the table top. As the mass moves up and down, its height \(h\), above the table top, is approximated by a sinusoidal function of the elapsed time \(t\), for a short period of time. Determine which of the following is the cosine equation that gives the height, \(h\), as a function of time, \(t\).
The minimum height is \(10\text{ cm}\), and the maximum height is \(50\text{ cm}\). The amplitude is \(\frac{\text{max}-\text{min}}{2}\). \(\text{Amplitude}=\frac{50-10}{2}=20\) The vertical displacement is \(\frac{\text{max}+\text{min}}{2}\). \(d=\frac{50+10}{2}=30\) The time from minimum to maximum is \(0.8\) seconds, which is half a period. \(\text{Period}=2(0.8)=1.6\) Thus \(b=\frac{2\pi}{1.6}\). Since the mass starts at the minimum height at \(t=0\), use a negative cosine model. Therefore \(h=-20\cos\left(\frac{2\pi t}{1.6}\right)+30\). Answer: A
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Item 7. Determine all values of \(x\), on the interval \(0\leq x
The expression \(\frac{1}{4}\tan x\) is undefined where \(\tan x\) is undefined. The tangent function is undefined at its vertical asymptotes. For \(0\leq x<2\pi\), \(\tan x\) is undefined at: \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). Multiplying by \(\frac{1}{4}\) does not change where the function is undefined. Therefore the expression is undefined at \(x=\frac{\pi}{2},\frac{3\pi}{2}\). Answer: B
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