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# What is the vertex of y=-(x + 2)^2 - 3x+9?

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Mark D. Share
Jun 19, 2018

get the equation into the standard form of a quadratic

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

Expand the brackets

$y = - \left({x}^{2} + 4 x + 4\right) - 3 x + 9$

Remove the brackets

$y = - {x}^{2} - 4 x - 4 - 3 x + 9$

Collect like terms

$y = - {x}^{2} - 7 x + 5$

Now use $\frac{- b}{2 a}$ to find the x coordinate of the vertex.

$\frac{- - 7}{2 \times - 1} = \frac{7}{- 2}$

Put this into the equation

$y = - {\left(\frac{7}{- 2}\right)}^{2} - 7 \times \frac{7}{- 2} + 5$

$y = - \frac{49}{4} + \frac{49}{2} + 5$

$y = \frac{69}{4}$

The maximum is $\left(- \frac{7}{2} , \frac{69}{4}\right)$

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