# What is the vertex of  y= (x+8)^2-2?

Feb 14, 2016

vertex$\to \left(x , y\right) \to \left(- 8 , - 2\right)$

#### Explanation:

When a quadratic is in this from ${x}_{\text{vertex}} = \left(- 1\right) \times b$

where $b \to {\left(x + b\right)}^{2}$

In truth, if the original equation was of form:
$y = a {x}^{2} + b + c$..............................(1)

and $k$ is a corrective value and you write equation (1) as:

$y = a {\left(x + \frac{b}{a}\right)}^{2} + k + c$

Then ${x}_{\text{vertex}} = \left(- 1\right) \times \frac{b}{a}$

However, in your case, $a = 1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${x}_{\text{vertex}} = \left(- 1\right) \times 8 = - 8$

Having found this just substitute into the original equation to find the value of ${y}_{\text{vertex}}$

So we have: $y = {\left(\left(- 8\right) + 8\right)}^{2} - 2 \text{ "=" } - 2$

so the vertex$\to \left(x , y\right) \to \left(- 8 , - 2\right)$

Feb 14, 2016

( -8 , -2 )

#### Explanation:

The equation of a parabola in vertex form is :

$y = {\left(x - h\right)}^{2} + k$
where (h , k ) are the coords of the vertex.

here $y = {\left(x + 8\right)}^{2} - 2$

and by comparison h = -8 and k = -2 → vertex = (-8 , -2 )
graph{(x+8)^2-2 [-10, 10, -5, 5]}