Question

What is the maximum for the parabola `y=3x^2-12x+8`?

Answer 1

Maximum is `oo` and minimum is `-4`.

Explanation:

As `y= graph{3x^2-12x+8 [-7.375, 12.625, -6.6, 3.4]}`

`=3(x^2-4x)+8`

`=3(x^2-4x+4)+8-12`

`=3(x-2)^2-4`

As `(x-2)^2>=0` we have a minimum value of `y` as `-4` at `x=2`

and there is no maxima as `y` can go to `oo`.