Answer
Please see the graph.
Work Step by Step
$f(x) = 2(x+3)^2-4$
$f(x) = 2(x-{-3})^2-4$
The axis of symmetry for a graph of the form $a(x-h)^2+k$ is $x=h$. So, the axis of symmetry for the line above is $x=-3$.
The vertex is as follows:
$f(x) = 2(x+3)^2-4$
$f(-3) = 2(-3+3)^2-4$
$f(-3) = 2(0)^2-4$
$f(-3) = 2*0-4$
$f(-3)=0-4$
$f(-3)=-4$
$(-3, -4)$ is the vertex.
$x=-5$
$f(x) = 2(x+3)^2-4$
$f(-5) = 2(-5+3)^2-4$
$f(-5) = 2(-2)^2-4$
$f(-5)=2*4-4$
$f(-5)=8-4$
$f(-5)=4$
Since we are two units from the axis of symmetry, we also have $(-1, 4)$ as another point on the curve.
$x=-4$
$f(x) = 2(x+3)^2-4$
$f(-4) = 2(-4+3)^2-4$
$f(-4) = 2(-1)^2-4$
$f(-4) = 2*1-4$
$f(-4) = 2-4$
$f(-4)=-2$
Since we are one unit from the axis of symmetry, we also have $(-2, -2)$ as another point on the curve.
