Question

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 1

The coordinates of the focus of the parabola are (-6,6)

Explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex

(h,k+(1/4a)) is the focus

in this problem we have

`y=-(1)/(12)x^(2)-x+6`

Convert to vertex form

`y-6=-(1)/(12)x^(2)-x`

`y-6=-(1)/(12)(x^(2)+12x)`

`y-6-3=-(1)/(12)(x^(2)+12x+36)`

`y-9=-(1)/(12)(x+6)^(2)`

`y=-(1)/(12)(x+6)^(2)+9` ------> equation in vertex form

The vertex is the point (-6,9)

`a=-(1)/(12)`

The focus is (h,k+(1/4a)) ------> (-6,9+(1/4(-1/12))-----> (-6,9-3)----> (-6,6)