Question

How do you find the maximum value of `y = -x^2 + 8x - 4`?

Answer 1

12

Explanation:

`y = -x^2 + 8x -4`

`:. y' = -2x + 8`

and `y''=-2` (See Note below)

As such we have maximum (as `y''<0`) at `y'=0 => -2x + 8 = 0`

`:. 2x = 8`
`:. x = 4`

When `x=4 => y=-4^2+8(4)-4 = -16+32-4 =12`

Note: As this is a quadratic and the coefficient of `x^2 <0` it should be obvious that the critical point corresponds to a maximum without the need to examine the second derivative.

Incidentally, this can also be fond without calculus by completing the square:
`y = -x^2 + 8x -4`
`y = -(x^2 - 8x +4)`
`y = -((x-4)^2 -4^2+4)`
`y = -((x-4)^2 -16+4)`
`y = -((x-4)^2 -12)`
`y = -(x-4)^2 +12`

So maximum of `y=12` occurs when `x-4=0=>x=4`
graph{-x^2 + 8x -4 [-16.17, 23.83, -4.8, 15.2]}