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

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Answer 1 · 2017-04-23T12:37:21

The vertex is $(- 1, - 2)$ $(-1,-2)$

To find the x coordinate, h, of the vertex, use the equation:

$h = - \frac{b}{2 (a)}$ $h = -b/(2(a))$ :

$h = - \frac{- 2}{2 (- 1)}$ $h = -(-2)/(2(-1))$

$h = - 1$ $h = -1$

To find the y coordinate, k, of the vertex, evaluate the function at $x = h$ $x = h$ :

$k = y (h)$ $k = y(h)$

$k = y (- 1)$ $k = y(-1)$

$k = - {(- 1)}^{2} - 2 (- 1) - 3$ $k = -(-1)^2-2(-1)-3$

$k = - 1 + 2 - 3$ $k = -1+2-3$

$k = - 2$ $k = -2$

The vertex is $(- 1, - 2)$ $(-1,-2)$

What is the vertex of y= -x^2 - 2x - 3 y=−x2−2x−3 y= -x^2 - 2x - 3 ?