# How do you find the vertex and the intercepts for y= x^2-8x+5?

Jul 21, 2018

$\text{vertex } = \left(4 , - 11\right)$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

•color(white)(x)y=a(x-h)^2+k

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to obtain this form "color(blue)"complete the square}$

$y = {x}^{2} + 2 \left(- 4\right) x + 16 - 16 + 5$

$\textcolor{w h i t e}{y} = {\left(x - 4\right)}^{2} - 11 \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(4 , - 11\right)$

$\text{to obtain the y-intercept let x = 0}$

$y = 0 - 0 + 5 = 5 \leftarrow \textcolor{red}{\text{y-intercept}}$

$\text{to obtain the x-intercepts let y = 0}$

${\left(x - 4\right)}^{2} - 11 = 0$

${\left(x - 4\right)}^{2} = 11$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$x - 4 = \pm \sqrt{11} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\text{add 4 to both sides}$

$x = 4 \pm \sqrt{11} \leftarrow \textcolor{red}{\text{exact solutions}}$

$x \approx 0.68 \text{ or "x~~7.32" to 2 dec. places}$