Answer: The period is \(2\pi\). The graph meets the axes at \((0,-3)\) and \(\left(\frac{\pi}{2},0\right)\).

For a trigonometric graph, identify the amplitude, midline and period first, then use these to locate the key points of the sketch.

(a) The period of \(\tan u\) is \(\pi\). Here \(u=\frac{\theta}{2}\), so the period in \(\theta\) is

\(2\pi.\)

(b) The \(y\)-intercept is found by putting \(\theta=0\):

\(y=3\tan0-3=-3.\)

So the graph meets the \(y\)-axis at

\((0,-3).\)

For the \(x\)-intercept, set \(y=0\):

\(3\tan\frac{\theta}{2}-3=0.\)

Hence

\(\tan\frac{\theta}{2}=1.\)

In the given interval, this gives

\(\frac{\theta}{2}=\frac{\pi}{4},\)

so

\(\theta=\frac{\pi}{2}.\)

Therefore the graph meets the \(x\)-axis at

\(\left(\frac{\pi}{2},0\right).\)

The graph is an increasing tangent curve on \(-\pi\lt\theta\lt\pi\), with vertical asymptotes at \(\theta=-\pi\) and \(\theta=\pi\).