# How do you find the vertex of a quadratic equation y=(x+8)^2-2?

Nov 28, 2015

${x}_{\text{vertex}} = - 8$

I will let you find ${y}_{\text{vetex}}$ by sustitution

#### Explanation:

Expanding the brackets gives:
$y = {x}^{2} + 16 x + 64 - 2$
$y = {x}^{2} + 16 x + 64$

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The equation given in the question is a quadratic that is already in vertex form (completing the square).

Consider the +8 from ${\left(x + 8\right)}^{2}$

The x_("vertex") " should be at "(-1)xx8=-8
Lets have a look!

Looks good!

Now all you have to do is substitute $x = \left(- 8\right)$ in the equation to find the value of y.

I will let you do that!