Answer: \(x=-\dfrac{5\pi}{12},-\dfrac{\pi}{12},\dfrac{\pi}{4}\). The graph has \(x\)-intercepts at these values and \(y\)-intercept \((0,4)\).

(a) Since

\(\tan3x=-1,\)

we can write

\(3x=-\frac{\pi}{4}+k\pi,\)

where \(k\) is an integer.

Since

\(-\frac{\pi}{2}\leq x\leq\frac{\pi}{2},\)

we need

\(-\frac{3\pi}{2}\leq3x\leq\frac{3\pi}{2}.\)

The values of \(3x\) in this interval are

\(-\frac{5\pi}{4},\qquad -\frac{\pi}{4},\qquad \frac{3\pi}{4}.\)

Therefore

\(x=-\frac{5\pi}{12},\qquad x=-\frac{\pi}{12},\qquad x=\frac{\pi}{4}.\)

(b) The graph is

\(y=4\tan3x+4.\)

It meets the \(x\)-axis when

\(4\tan3x+4=0,\)

which gives

\(\tan3x=-1.\)

So the \(x\)-intercepts are

\(\left(-\frac{5\pi}{12},0\right),\quad \left(-\frac{\pi}{12},0\right),\quad \left(\frac{\pi}{4},0\right).\)

The \(y\)-intercept occurs when \(x=0\):

\(y=4\tan0+4=4.\)

So the \(y\)-intercept is

\((0,4).\)

The vertical asymptotes occur when \(3x=\frac{\pi}{2}+k\pi\), so within the interval they occur at

\(x=-\frac{\pi}{2},\quad -\frac{\pi}{6},\quad \frac{\pi}{6},\quad \frac{\pi}{2}.\)

The sketch should therefore have three increasing tangent branches between these asymptotes, passing through the three \(x\)-intercepts and the point \((0,4)\).