Answer
$$
y=-2x^{2}-12x-16
$$
The $x$-intercepts are $-2 , -4 $.
The $y$-intercept is -16.
Vertex parabola is $\left(-3 , 2 \right)$.
The axis is $x=-3 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-3 , 2 \right)$ gives the graph in Figure
Work Step by Step
$$
y=-2x^{2}-12x-16
$$
The $x$-intercepts can be found by letting $y=0$ to get
$$
y=-2x^{2}-12x-16=-2(x^{2}+6x+8)=-2(x+2)(x+4)=0
$$
from which $x=-2 $ and $x=-4 $ are the $x$-intercepts.
Set $x=0 $ to find the $y$-intercept.
$$
y=-2(0)^{2}-12.(0)-16=-16.
$$
So the $y$-intercept is -16.
The $x$-coordinate of the vertex is :
$$
x=\frac{-b}{2a}=\frac{12}{-4}=-3
$$
Substituting this into the equation gives
$$
y=-2(-3)^{2}-12(-3)-16=-18+36-16=2
$$
The vertex is $\left(-3 , 2 \right)$.
The axis is $x=-3 $ , the vertical line through the vertex.
Plotting the vertex, the $y$-intercept,the $x$-intercepts. and the point $\left(-3 , 2\right)$ gives the graph in Figure