The vertex form of a parabola is:
[tex]y=a(x-h)^2+k[/tex]where a is a coefficient, and (h,k) is the vertex.
Substituting with (0, 6) as the vertex, we get:
[tex]\begin{gathered} y=a(x-0)^2+6 \\ y=ax^2+6 \end{gathered}[/tex]Substituting with the point (-8, -10), we can find a, as follows:
[tex]\begin{gathered} -10=a(-8)^2+6 \\ -10=a\cdot64^{}+6 \\ -10-6=a\cdot64 \\ \frac{-16}{64}=a \\ -\frac{1}{4}=a \end{gathered}[/tex]Therefore, the equation is:
[tex]y=-\frac{1}{4}x^2+6[/tex]