Answer
$\frac{9}{4}$
Work Step by Step
Consider a point $(x,x+2)$ on the line $y=x+2$ and the corresponding point having the same $x$-coordinate $(x,x^2)$ on the parabola $y=x^2$.
Let the vertical distance be given by
$v(x)$ = $(x+2)-x^{2}$, $-1$ $\leq$ $x$ $\leq$ $2$
$v'(x)$ = $1-2x$ = $0\Rightarrow x$ = $\frac{1}{2}$
$v(-1)$ = $0$
$v\left(\frac{1}{2}\right)$ = $\frac{9}{4}$
$v(2)$ = $0$
So there is an absolute maximum at $x$ = $\frac{1}{2}$.
The maximum distance is $\frac{9}{4}$.
