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

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Answer 1 · 2018-06-19T18:32:57

get the equation into the standard form of a quadratic

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

Expand the brackets

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

Remove the brackets

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

Collect like terms

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

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

$\frac{- - 7}{2 \times - 1} = \frac{7}{- 2}$ $(- -7)/(2xx -1)=7/(-2)$

Put this into the equation

$y = - {(\frac{7}{- 2})}^{2} - 7 \times \frac{7}{- 2} + 5$ $y=-(7/(-2))^2-7xx7/(-2)+5$

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

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

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

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