What are the coordinates of the focus of the parabola?y= −1/12x^2−x+6(−6, 6)(6, 6)(6, 9)(−6, 9)

Answer:The coordinates of the focus of the parabola are (-6,6)Step-by-step explanation:we know thatThe 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 focusin 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 formThe 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)