Question

graph the trigonometric function.

$y = \frac{1}{2} \cos x$

plot all points corresponding to x-intercepts, minima, and maxima within one cycle. then click on the graph-a-function button.

Explanation:

Step1: Identify the parent function and transformations

The parent function is \( y = \cos x \), which has a period of \( 2\pi \), amplitude \( 1 \), maximum value \( 1 \), minimum value \( -1 \), \( x \)-intercepts at \( x=\frac{\pi}{2}, \frac{3\pi}{2} \) within \( [0, 2\pi] \), maximum at \( x = 0, 2\pi \) (value \( 1 \)), minimum at \( x=\pi \) (value \( -1 \)). For \( y=\frac{1}{2}\cos x \), the amplitude is \( \frac{1}{2} \), so the maximum value is \( \frac{1}{2} \), minimum is \( -\frac{1}{2} \), period remains \( 2\pi \), \( x \)-intercepts same \( x=\frac{\pi}{2}, \frac{3\pi}{2} \), maxima at \( x = 0, 2\pi \) (value \( \frac{1}{2} \)), minimum at \( x=\pi \) (value \( -\frac{1}{2} \)).

Step2: Find key points within one cycle (\( [0, 2\pi] \))

• Maxima: When \( \cos x = 1 \) (i.e., \( x = 0, 2\pi \)), \( y=\frac{1}{2}(1)=\frac{1}{2} \). So points \( (0, \frac{1}{2}) \) and \( (2\pi, \frac{1}{2}) \).
• Minimum: When \( \cos x = -1 \) (i.e., \( x = \pi \)), \( y=\frac{1}{2}(-1)=-\frac{1}{2} \). So point \( (\pi, -\frac{1}{2}) \).
• x-intercepts: When \( \cos x = 0 \) (i.e., \( x=\frac{\pi}{2}, \frac{3\pi}{2} \)), \( y = 0 \). So points \( (\frac{\pi}{2}, 0) \) and \( (\frac{3\pi}{2}, 0) \).

Answer:

To graph \( y = \frac{1}{2}\cos x \) within one cycle (\( 0 \leq x \leq 2\pi \)):

• Plot maxima at \( (0, \frac{1}{2}) \) and \( (2\pi, \frac{1}{2}) \),
• Plot minimum at \( (\pi, -\frac{1}{2}) \),
• Plot x-intercepts at \( (\frac{\pi}{2}, 0) \) and \( (\frac{3\pi}{2}, 0) \),

then connect these points smoothly to form the cosine curve with amplitude \( \frac{1}{2} \).