How do you find the vertex of this parabola $y=-x^2-2x-5$ ?

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Answer 1 · 2016-11-03T19:51:33

Please read the explanation for the steps leading to the vertex $(1, -8)$

For an equation of a parabola of the form $y = ax^2 + bx + c$ , the x coordinate, h, of the vertex is:

$h = -b/(2a)$

In the given equation, $a = -1, and b = -2$ , therefore:

$h = (-2)/(2(-1))$

$h = 1$

To obtain the y coordinate of the vertex, evaluate the function at h:

$y = -(1)^2 - 2(1) - 5$

$y = -8$

The vertex is at the point $(1, -8)$

How do you find the vertex of this parabola y=-x^2-2x-5y=-x^2-2x-5?