Answer
See below
Work Step by Step
The graphs intersect where the points on the two graphs coincide.
Obtain: $\frac{1}{2}x^2-3x+4=2x^2-5x-12\\
\Leftrightarrow 2x^2-5x-12-\frac{1}{2}x^2+3x-4=0\\
\Leftrightarrow \frac{3}{2}x^2-2x-16=0\\
\Leftrightarrow 3x^2-4x-32=0$
The solution to this: $x=\frac{4 \pm \sqrt 400}{6}\\
\rightarrow x=\frac{4+20}{6} \lor x=\frac{4-20}{6}\\
\rightarrow x=4 \lor x=-\frac{8}{3}$
To find the $y$ value:
$f(4)=0\\f(-\frac{8}{3})=\frac{140}{9}$
Hence, we have two points $(4,0)$ and $(-\frac{8}{3},\frac{140}{9})$.
