# What is the axis of symmetry and vertex for the graph y=-x^2+x+12?

Mar 5, 2016

$\textcolor{b l u e}{\text{Axis of symmetry } \to x = \frac{1}{2}}$

color(green)("Vertex "->" "(x,y)" "->" " (1/2,12 1/4)

#### Explanation:

It is not uncommon for people to be shown the method of completing the square to solve this context. At first it is quite confusing so I am going to show you something that is part way towards completing the square as an alternative.

Given:$\text{ } y = - {x}^{2} + x + 12$

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Compare to $y = a {x}^{2} + b x + c$

Rewritten as:$\text{ } a \left({x}^{2} + \frac{b}{a} x\right) + c$

Then you have:" "x_("vertex")=(-1/2)xxb/a
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$a = \left(- 1\right)$
$b = \left(+ 1\right)$

So we have:

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{1}{- 1} = + \frac{1}{2}}$
$\textcolor{b l u e}{\text{Axis of symmetry } \to x = \frac{1}{2}}$
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Substitute $x = \frac{1}{2}$ in the original equation and you should end up with:

$\textcolor{b l u e}{{y}_{\text{vertex}} = 12 \frac{1}{4}}$

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color(green)("Vertex "->" "(x,y)" "->" " (1/2,12 1/4)