V(-4,3) point (-6,11)
There is more than one parabola that fits this description - an infinite number I think. You need 3 points for a unique parabola. However if the axis of symmetry is parallel to the y axis then there is only one and this is a fair assumption for your question.
Let's look at your answer:
1. v(-4,3); pt (-6,11)
y=a(x-h)^2+k alright
y=a(x--4)^2+3 added
y=a(x+4)^2+3 added
The h and k come directly from the vertex. You sub those in first and then use the other point to determine the value of a.
11= a(-6-(-3)+(-3) incorrect (I won't worry about the rest)
I'll redo this for you (your method)
$$v(-4,3); pt (-6,11)\\
\begin{array}{rlll}
\mbox{First sub in the vertex}\\
y&=&a(x-h)^2+k \\
y&=&a(x--4)^2+3 \\
y&=&a(x+4)^2+3 \\
\mbox{Now sub in the other point to find a}\\
11&=&a(-6+4)^2+3 \\
8&=&a(-2)^2 \\
8&=&4a \\
a&=&2 \\
\mbox{Therefore}\\
y&=&2(x+4)^2+3 &\mbox{This is the vertex form}\\
\end{array}$$
$$\mbox{Now I will convert it to standard form }\\\\
\begin{array}{rlll}
(x+4)^2&=&\frac{1}{2} (y-3)\\
x^2+8x+16&=&\frac{1}{2} (y-3)\\
2x^2+16x+32&=& (y-3)\\
2x^2+16x+35&=& y\\
y&=& 2x^2+16x+35\\
\end{array}\\\\$$
(I have edited this to get rid of irrelevant calculations)
