# Does the function f(x)=-8x +x^2 have a maximum or minimum?

Aug 3, 2017

There is a minimum at $\left(4 , - 16\right)$

#### Explanation:

We need

$\left({x}^{n}\right) ' = n {x}^{n - 1}$

Our function is

$f \left(x\right) = {x}^{2} - 8 x$

We calculate the first derivative,

$f ' \left(x\right) = 2 x - 8$

The critical point is when $f ' \left(x\right) = 0$

$2 x - 8 = 0$, $\implies$, $x = 4$

The second derivative is

$f ' ' \left(x\right) = 2$

As

$f ' ' \left(x\right) > 0$, the function is convex at $x = 4$

graph{x^2-8x [-31.13, 33.82, -22.35, 10.13]}