# What is the axis of symmetry and vertex for the graph  y-8=-2(x-3)^2?

Jan 9, 2017

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#### Explanation:

$\textcolor{b r o w n}{\text{There is a shortcut to this that is part of completing the square}}$

You need the form of $y = a {x}^{2} + b x + c$

x_("vertex")=(-1/2)xxb/a -> "axis of symmetry"

Given:$\text{ } y - 8 = - 2 {\left(x - 3\right)}^{2}$

$\implies y = - 2 \left({x}^{2} - 6 x + 9\right) + 8$

$\implies y = - 2 {x}^{2} + 12 x - 10$

so ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{12}{- 2} = + 3$