Answer
General shape:
Work Step by Step
$y'=\tan x$
$\left[\begin{array}{llllll}
y': & & -- & | & ++ & \\
& (-\frac{\pi}{2} & & 0 & & \frac{\pi}{2})\\
y: & & \searrow & \min & \nearrow &
\end{array}\right]$
$y''=\sec^{2}x,$
is never negative on $(-\displaystyle \frac{\pi}{2},\frac{\pi}{2})$, so $y$ is concave up.
(no points of inflection)
The graph falls from $+\infty$ at the left end of the interval $(x=-\displaystyle \frac{\pi}{2})$ to $x=0$, from where it rises without bound at the right end of the interval $(x=\displaystyle \frac{\pi}{2})$.
