The equation of the parabola in vertex form is
[tex]y=a(x-h)^2+k[/tex]where the point (h,k) is the coordinate of the vertex. From our picture, we can note that (h,k)=(-6,-4).
By substituting these values into our first equation, we have
[tex]y=a(x-(-6))^2-4[/tex]which gives
[tex]y=a(x+6)^2-4[/tex]Now, we can find the constant a by substituting one of the other given point. If we choose point (0,-2) into this last equation, we get
[tex]-2=a(0+6)^2-4[/tex]which gives
[tex]\begin{gathered} -2=a(6^2)-4 \\ -2=36a-4 \end{gathered}[/tex]then, by moving -4 to the left hand side, we have
[tex]\begin{gathered} -2+4=36a \\ 2=36a \\ or\text{ equivalently,} \\ 36a=2 \end{gathered}[/tex]and finally, a is equal to
[tex]\begin{gathered} a=\frac{2}{36} \\ a=\frac{1}{18} \end{gathered}[/tex]hence, the equation of the parabola in vertex form is
[tex]y=\frac{1}{18}(x+6)^2-4[/tex]Now, lets convert this equation into a standrd form. This can be done by expanding the quadratic term and collecting similar term. That is, by expanding the quadratic terms, we obtain
[tex]y=\frac{1}{18}(x^2+12x+36)-4[/tex]now, by distributing 1/18, we have
[tex]y=\frac{1}{18}x^2+\frac{12}{18}x+\frac{36}{18}-4[/tex]which is equivalent to
[tex]y=\frac{1}{18}x^2+\frac{1}{3}x+2-4[/tex]and finally, the parabola equation in standard form is
[tex]y=\frac{1}{18}x^2+\frac{1}{3}x-2[/tex]