Answer
The graph of $y=2x^2+4x+8$ has no $x-$intercepts.
Work Step by Step
The given function is
$\Rightarrow y=2x^2+4x+8$
Find the number of real solutions of $0=2x^2+4x+8$
$= b^2-4ac$
Substitute $2$ for $a,4$ for $b,$ and $8$ for $c$.
$= (4)^2-4(2)(8)$
Simplify.
$= 16-64$
Subtract.
$= -48$
Discriminant is less than $0$,
Hence, the equation has no real solutions.
The graph of $y=2x^2+4x+8$ has no $x-$intercepts.
