# What are the values and types of the critical points, if any, of f(x)=x^2-5x+4?

A Minimum point $\left(\frac{5}{2} , - \frac{9}{4}\right)$

#### Explanation:

$f \left(x\right) = {x}^{2} - 5 x + 4$

First derivative $f ' \left(x\right) = 2 x - 5$

Equate the $f ' \left(x\right) = 0$

$2 x - 5 = 0$

solve for $x$

$x = \frac{5}{2}$

solve for $y = {x}^{2} - 5 \left(x\right) + 4$

$y = {\left(\frac{5}{2}\right)}^{2} - 5 \left(\frac{5}{2}\right) + 4$

$y = \frac{25}{4} - \frac{25}{2} + 4$

$y = \frac{25 - 50 + 16}{4}$

$y = - \frac{9}{4}$