# How do you find the vertex and the intercepts for f(x)=-x^2+2x+4?

Jul 30, 2018

$\text{vertex } = \left(1 , 5\right) , x = 1 \pm \sqrt{5}$

#### Explanation:

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

•color(white)(x)y=a(x-h)^2+k

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

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

$y = 1 \left({x}^{2} + 2 \left(- 1\right) x + 1 - 1 - 4\right)$

$\textcolor{w h i t e}{y} = - {\left(x - 1\right)}^{2} + 5$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(1 , 5\right)$

$\text{let x = 0 for y-intercept}$

$y = - 1 + 5 = 4 \leftarrow \textcolor{red}{\text{y-intercept}}$

$\text{let y = 0 for x-intercepts}$

$- {\left(x - 1\right)}^{2} + 5 = 0$

${\left(x - 1\right)}^{2} = 5$

$x - 1 = \pm \sqrt{5}$

$x = 1 \pm \sqrt{5} \leftarrow \textcolor{red}{\text{exact values}}$