Question

Question #9ea92

Answer 1

The vertex form of a parabola is
`y =m(x-a)^2+b`

Given `y=3x^2+18x-5`

`y = 3(x^2+6x) -5`

`y = 3(x^2+6x+9) -5 -27`

`y=3(x+3)^2 - 32`

`y= 3(x-(-3))^2+(-32)`
-that is the vertex is at `(-3,-32)`