Question

Question #eb9ad

Answer 1

`y = -8x^2-32x-12`

Explanation:

`y = ax^2 + bx +c`
`y = p(x+q)^2+r`

`y-`intercept `=(0,-12) -> c=12`

`ax^2 + bx +c = ax^2+bx-12`

vertex `= (-q,r)`

`(-q,r) = (-2,20) -> q=2, r=20`

this means that `y = p(x+2)^2+20`

expanding this gives `y = px^2+4px+4p+20`

`ax^2 + bx +c = px^2 + 4px + (4p+20)`

`c = 4p + 20`

(we are given that the `y-`intercept is `(0,-12)`)

`c = -12 -> 4p+20 = -12`

`4p = -32`

`p = -8`

inputting `p=8`:

`y = -8(x+2)^2+20`

`y = -8(x^2+4x+4) + 20`
`y = -8x^2-32x-32 + 20`
`y = -8x^2-32x-12`

graph{-8x^2-32x-12 [-22.42, 17.58, 3.76, 23.76]}
(graph with equation `y=-8x^2-32x-12`)
vertex: `(-2,20)`
`y-`intercept: `(0,-12)`