ANSWER
[tex]y=-2(x+1)^2-4[/tex]EXPLANATION
The vertex form of a quadratic equation is:
[tex]y=a(x-h)^2+k[/tex]where (h, k) = vertex
Let us substitute the given vertex and the given point into the equation to find the value of the constant a.
That is:
[tex]\begin{gathered} -36=a(-5-(-1))^2+(-4) \\ -36=a(-5+1)^2-4 \\ -36=a(-4)^2-4 \\ -36=16a-4 \\ 16a=-36+4=-32 \\ \Rightarrow a=\frac{-32}{16} \\ a=-2 \end{gathered}[/tex]That is the value of the constant a. Now we can write the quadratic function using the constant a and the vertex:
[tex]\begin{gathered} y=-2(x-(-1))^2+(-4) \\ y=-2(x+1)^2-4 \end{gathered}[/tex]That is the function.