# How do you find the vertex and intercepts for y=x^2+4x+1?

Oct 10, 2017

$\left(- 2 , - 3\right) , x = - 2 \pm \sqrt{3}$

#### Explanation:

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

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

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

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

• " ensure the coefficient of "x^2" term is 1"

• " add/subtract "(1/2"coefficient of x-term")^2" to "x^2+4x

$y = {x}^{2} + 4 x + 4 \leftarrow \text{ coefficient of "x^2" term is 1}$

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

$\textcolor{w h i t e}{y} = {\left(x + 2\right)}^{2} - 3$

$\Rightarrow \text{vertex } = \left(- 2 , - 3\right)$

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

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

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

$x = 0 \to y = 4 - 3 = 1 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to {\left(x + 2\right)}^{2} - 3 = 0$

$\Rightarrow {\left(x + 2\right)}^{2} = 3$

$\Rightarrow x + 2 = \pm \sqrt{3} \leftarrow \text{ note plus or minus}$

$\Rightarrow x = - 2 \pm \sqrt{3} \leftarrow \textcolor{b l u e}{\text{exact values}}$

$x \approx - 3.73 , x \approx - 0.27 \leftarrow \textcolor{red}{\text{ x-intercepts}}$