[tex]\bf (y-k)=a(x-h)^2\qquad
\begin{cases}
vertex\ (h,k)\\
vertex\ (-4,2)
\end{cases}\\\\\\ y-2=a(x-(-4))^2
\implies
y=a(x+4)^2+2[/tex]
so.. what is the coefficient "a"?
well
now, we know, another point on the graph, besides the vertex, we know a y-intercept, that is, 0, -30, that simply means when x = 0, y = -30
[tex]\bf y=a(x+4)^2+2\qquad (0,-30)\implies -30=a(0+4)^2+2
\\\\\\
-30-2=a4^2\implies \cfrac{-32}{16}=a\implies -2=a
\\\\\\
thus\implies y=-2(x+4)^2+2[/tex]
now, getting the x-intercepts, is just the zeros, or solution to the quadratic
[tex]\bf y=-2(x+4)^2+2\impliedby \textit{setting y to 0}
\\\\\\
0=-2(x+4)^2+2\implies -2=-2(x+4)^2\implies \cfrac{-2}{-2}=(x+4)^2
\\\\\\
1=(x+4)^2\implies \pm\sqrt{1}=x+4\implies \pm 1-4 = x\to
\begin{cases}
(-3\ ,\ 0)\\
(-5\ ,\ 0)
\end{cases}[/tex]
notice, "y" is 0 on both cases, because, is an x-intercept, or a zero, and when the graph touches the x-axis, "y" is zero
