Answer
$$
y=2x^{2}+8x-8
$$
The $x$-intercepts are
$$
x_{1}=2\left(\sqrt{2}-1\right) \approx 0.828 ,\:x_{2}=-2\left(1+\sqrt{2}\right) \approx -4.828
$$
The $y$-intercept is -8.
Vertex parabola is $\left(-2 , -16 \right)$.
The axis is $x=-2$ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-2 , -16 \right)$. gives the graph in Figure
Work Step by Step
$$
y=2x^{2}+8x-8
$$
The $x$-intercepts can be found by letting $y=0$ to get
$$
y=2x^{2}+8x-8=x^{2}+4x-4=0
$$
This does not appear to factor, so we’ll try the quadratic formula.
$$
\:x_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:1\left(-4\right)}}{2\cdot \:1}
$$
from which
$$
x_{1}=2\left(\sqrt{2}-1\right) \approx 0.828 ,\:x_{2}=-2\left(1+\sqrt{2}\right) \approx -4.828
$$
are the $x$-intercepts.
To find the $y$-intercept , set $x=0 $
$$
y=2. (0)^{2}+8(0)-8=-8.
$$
So the $y$-intercept is -8.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=-\frac{8}{4}=-2
$$
Substituting this into the equation gives
$$
y=2(-2)^{2}+8.(-2)-8=8-16-8=-16
$$
The vertex is $\left(-2 , -16 \right)$.
The axis is $x=-2 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-2 , -16 \right)$ gives the graph in Figure