Question

How do you find the local max and min for `f(x)= x^2 + 8x -12`?

Answer 1

`y_{min} = -28` at `x = -4`

Explanation:

calculus is OTT for this

So

`y = x^2 + 8x - 12 = (x+4)^2 - 16 - 12`
`= (x+4)^2 - 28`

`(x+4)^2 \ge 0` so `y_{min} = -28` at `x = -4`