# Does f(x)=2x^2-8x+5 have a maximum or a minimum? If so, what is it?

Dec 6, 2016

$f \left(x\right) = 2 {x}^{2} - 8 x + 5$ has a minimum value of $\left(- 3\right)$ at $x = 2$

#### Explanation:

A parabola with expressed in the form: $\textcolor{red}{a} {x}^{2} + b x = c$
has a maximum if $\textcolor{red}{a} < 0$
or a minimum if $\textcolor{red}{a} > 0$

$f \left(x\right) = \textcolor{red}{2} {x}^{2} - 8 x + 5$ must, therefore, have a minimum.

The minimum will occur when the tangent slope is $0$;
or expressed in another way, when the derivative of $f \left(x\right)$ is equal to $0$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = 4 x - 8$

and $4 x - 8 = 0$
when $x = \textcolor{b l u e}{2}$

$f \left(x = \textcolor{b l u e}{2}\right) = 2 \cdot {\textcolor{b l u e}{2}}^{2} - 8 \cdot \textcolor{b l u e}{2} + 5$
$\textcolor{w h i t e}{\text{XX}} = 8 - 16 + 5$
$\textcolor{w h i t e}{\text{XX}} = - 3$