Answer
The graph of $y=-\frac{1}{2}x^2+2x$ has two $x-$intercepts.
Work Step by Step
The given function is
$\Rightarrow y=-\frac{1}{2}x^2+2x$
Find the number of real solutions of $0=-\frac{1}{2}x^2+2x$
$= b^2-4ac$
Substitute $-\frac{1}{2}$ for $a,2$ for $b,$ and $0$ for $c$.
$= (2)^2-4(-\frac{1}{2})(0)$
Simplify.
$= 4-0$
Subtract.
$= 4$
Discriminant is greater than $0$,
Hence, the equation has two real solutions.
The graph of $y=-\frac{1}{2}x^2+2x$ has two $x-$intercepts.
