Answer
amplitude=$2$
period =$\displaystyle \frac{2}{3}$
graph (black):
.
Work Step by Step
See p. 424
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: $b$
------------------
$y=-2\cos 3\pi x$
$a=-2 , k=4\pi, b=0$
So,
amplitude=$|-2|=2$
period =$\displaystyle \frac{2\pi}{3\pi}=\frac{2}{3}$
To graph,
begin with $f(x)=\cos x,$ (red, dashed)
horizontally compress by factor $ 3\pi$ and
reflect across the x-axis (black, solid line).
$g(0)=-2$
$g(\displaystyle \frac{1}{6})=-2\cos(\frac{\pi}{2})=0$
$g(\displaystyle \frac{1}{3})=-2\cos(\pi)=2$
$g(\displaystyle \frac{1}{2})=-2\cos(\frac{3\pi}{2})=0$
$g(\displaystyle \frac{2}{3})=-2\cos(2\pi)=-2.$