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

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Answer 1 · 2015-11-28T20:15:24

$x_("vertex")=-8$

I will let you find $y_("vetex")$ by sustitution

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

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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 $(x+8)^2$

The $x_("vertex") " should be at "(-1)xx8=-8$
Lets have a look!
Tony B
Looks good!

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

I will let you do that!

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