Answer:
The coordinates of the focus of the parabola are (-6,6)
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
(h,k+(1/4a)) is the focus
in this problem we have
[tex]y=-\frac{1}{12}x^{2}-x+6[/tex]
Convert to vertex form
[tex]y-6=-\frac{1}{12}x^{2}-x[/tex]
[tex]y-6=-\frac{1}{12}(x^{2}+12x)[/tex]
[tex]y-6-3=-\frac{1}{12}(x^{2}+12x+36)[/tex]
[tex]y-9=-\frac{1}{12}(x+6)^{2}[/tex]
[tex]y=-\frac{1}{12}(x+6)^{2}+9[/tex] ------> equation in vertex form
The vertex is the point (-6,9)
[tex]a=-\frac{1}{12}[/tex]
The focus is (h,k+(1/4a)) ------> (-6,9+(1/4(-1/12))-----> (-6,9-3)----> (-6,6)
