Answer
False.
The graph of $f(x)=-2(x+4)^2-8$ has one $y-$intercept and no $x-$intercepts.
Work Step by Step
The given function is a quadratic function:
$f(x)=-2(x+4)^2-8$.
For the $x-$intercepts.
Replace $f(x)$ with $0$ into the given function.
$\Rightarrow 0=-2(x+4)^2-8$
Add $8$ to both sides.
$\Rightarrow 0+8=-2(x+4)^2-8+8$
Simplify.
$\Rightarrow 8=-2(x+4)^2$
Divide both sides by $-2$.
$\Rightarrow \frac{8}{-2}=\frac{-2(x+4)^2}{-2}$
Simplify.
$\Rightarrow -4=(x+4)^2$
Take square root both sides.
$\Rightarrow \sqrt{-4}=\sqrt{(x+4)^2}$
The equation has imaginary solutions, there are no $x-$intercepts.
For the $y-$intercept.
Replace $x$ with $0$ in the given function.
$\Rightarrow f(0)=-2(0+4)^2-8$
Simplify.
$\Rightarrow f(0)=-2(4)^2-8$
$\Rightarrow f(0)=-32-8$
$\Rightarrow f(0)=-40$
The $y-$intercept is $-40$. The parabola passes through $(0,-40)$.
Hence, the given statement is false.
A true statement is:
The graph of $f(x)=-2(x+4)^2-8$ has one $y-$intercept and no $x-$intercepts.
