Question

Question #967d4

Answer 1

see explanation below

Explanation:

`Y = -x^2 + 4 x -12`

we can find it vertex by using completing a square.
`Y = -(x^2 - 4x) - 12`
`Y = -(x -2)^2 +(-2)^2 -12`
`Y = -(x -2)^2 + 4 -12`
`Y = -(x - 2)^2 - 8`
Therefore it vertex is maximum at `(2, -8)`

To find x-intercept, plug in `Y = 0`, but since it maximum is `-8` this equation has no x-intercept. We can check using `b^2-4ac < 0` where, `a =-1, b = 4 and c = -12`
(4)^2 -4(-1)(-12) = 16 -48 < 0

To find y-intercept, plug in `x = 0`,
`Y = -(0 - 2)^2 - 8`
`Y = -4 - 8 = -12` `->`y-intercept