# How do you find the vertex and the intercepts for f(x) =(x+5)^2 -1?

Nov 14, 2017

$\left(- 5 , - 1\right) \text{ and } x = - 6 , x = - 4$

#### Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

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

$f \left(x\right) = {\left(x + 5\right)}^{2} - 1 \text{ is in vertex form}$

$\text{with "h=-5" and } k = - 1$

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

$\textcolor{b l u e}{\text{Intercepts}}$

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

$x = 0 \to y = {\left(5\right)}^{2} - 1 = 24 \leftarrow \textcolor{red}{\text{y-intercept}}$

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

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

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

$\sqrt{{\left(x + 5\right)}^{2}} = \pm \sqrt{1} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x + 5 = \pm 1$

$\text{subtract 5 from both sides}$

$\Rightarrow x = - 5 \pm 1$

$x = - 5 - 1 = - 6 , x = - 5 + 1 = - 4 \leftarrow \textcolor{red}{\text{x-intercepts}}$