Answer
The graph of $f(x)=2x^2+8x+8$ has one $x-$intercept.
Work Step by Step
The given function is
$\Rightarrow f(x)=2x^2+8x+8$
Find the number of real solutions of $0=2x^2+8x+8$
$= b^2-4ac$
Substitute $2$ for $a,8$ for $b,$ and $8$ for $c$.
$= (8)^2-4(2)(8)$
Simplify.
$=64-64$
Subtract.
$= 0$
Discriminant is equal to $0$,
Hence, the equation has one real solutions.
The graph of $f(x)=2x^2+8x+8$ has one $x-$intercept.
