Answer
a.
amplitude=$\displaystyle \frac{1}{2}$,
period=$\pi$,
horizontal shift=$-\displaystyle \frac{\pi}{3}$
b.
$y=-\displaystyle \frac{1}{2}\cos[2(x+\frac{\pi}{3})]$
Work Step by Step
For
$y=a\sin k(x-b),\ ,\quad y=a\cos k(x-b)$,
$(k>0)$,
Amplitude: $|a|$,
Period: $\displaystyle \frac{2\pi}{k}$,
Horizontal shift: $b$
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If we reflect the graph over the x-axis, we see a cosine shape but the maximum is not at x=0. It is at $x=-\displaystyle \frac{\pi}{3}$.
So, there was a horizontal shift to the left, $b=-\displaystyle \frac{\pi}{3}.$
The amplitude is $\displaystyle \frac{1}{2}$, so a can be $\displaystyle \pm\frac{1}{2}$.
Since there is a reflection, a is negative.
That is, $a=-\displaystyle \frac{1}{2}$
The period (from minimum to minimum) here is
$\displaystyle \frac{2\pi}{3}-(-\frac{\pi}{3})=\pi$
From $\displaystyle \frac{2\pi}{k}=\pi$, it follows that k=2.
With these parameters,
$f(x)=y=a\cos k(x-b)$
$y=-\displaystyle \frac{1}{2}\cos[2(x-\frac{-\pi}{3})]$
$y=-\displaystyle \frac{1}{2}\cos[2(x+\frac{\pi}{3})]$
a.
amplitude=$\displaystyle \frac{1}{2}$,
period=$\pi$,
horizontal shift=$-\displaystyle \frac{\pi}{3}$
b.
$y=-\displaystyle \frac{1}{2}\cos[2(x+\frac{\pi}{3})]$
