# How do you find the vertex and intercepts for y = (x + 5)^2 – 2?

Jul 19, 2016

Vertex->(x,y)=(-5,-2)#

${y}_{\text{intercept}} \to y = 23$

${x}_{\text{intercept}}$ as follows:
$x = - 5 \pm \sqrt{2}$

$x \approx - 6.414$ to 3 decimal places
$x \approx - 3.586$ to 3 decimal places

#### Explanation:

$\textcolor{b l u e}{\text{Determine the vertex}}$

This is already in a vertex form the of quadratic equation so you can read off the coordinates of the vertex directly.

${x}_{\text{vertex}} = \left(- 1\right) \times \left(+ 5\right) = - 5$
${y}_{\text{vertex}} = - 2$
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$\textcolor{b l u e}{\text{Determine y intercept}}$

y intercept is at $x = 0$

$\implies y = {\left(0 + 5\right)}^{2} - 2 \to y = 23$
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$\textcolor{b l u e}{\text{Determine x intercept}}$

Set $y = 0$ giving

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

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

$\implies 2 = {\left(x + 5\right)}^{2} \text{ "vec("swap round") " } {\left(x + 5\right)}^{2} = 2$

Square root both sides

$x + 5 = \sqrt{2}$

Subtract 5 from both sides

$x = - 5 \pm \sqrt{2}$

$x \approx - 6.414$ to 3 decimal places
$x \approx - 3.586$ to 3 decimal places 