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Find vertex for f(x) = -x*2-2x-6. show work and sensable answer please.

2 Answers

3 votes


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ f(x)=\stackrel{\stackrel{a}{\downarrow }}{-1}x^2\stackrel{\stackrel{b}{\downarrow }}{-2}x\stackrel{\stackrel{c}{\downarrow }}{-6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{-2}{2(-1)}~~,~~-6-\cfrac{(-2)^2}{4(-1)} \right)\implies \left( \cfrac{2}{-2}~,~-6+\cfrac{4}{4} \right) \\\\\\ (-1~,~-6+1)\implies (-1~,~-5)

User Bacardi
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5.6k points
2 votes

Answer:

The vertex would be (-1, -5).

Explanation:

In order to find this, first find the x-coordinate of the vertex. You can do this by calculating out -b/2a in which a is the coefficient of x^2 and b is the coefficient of x.

-b/2a

-(-2)/2(-1)

2/-2

-1

So we know the x-coordinate to be -1. Now we plug that into the equation and find the y value.

-x^2 - 2x - 6

-(-1)^2 - 2(-1) - 6

-(1) + 2 - 6

-5

User Kishen
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5.1k points