# What is the vertex form of y= - x^2 - 17x - 15 ?

Feb 11, 2016

$y = - 1 {\left(x + \frac{17}{4}\right)}^{2} + 57 \frac{1}{4}$

#### Explanation:

Given -

$y = - {x}^{2} - 17 x - 15$

Find the vertex -

$x = \frac{- b}{2 a} = \frac{- \left(- 17\right)}{2 \times \left(- 1\right)} = \frac{17}{- 2} = \frac{- 17}{2}$

$y = - {\left(\frac{- 17}{2}\right)}^{2} - 17 \left(\frac{- 17}{2}\right) - 15$
$y = - \left(72 \frac{1}{4}\right) + 144 \frac{1}{2} - 15$
$y = - 72 \frac{1}{4} + 144 \frac{1}{2} - 15$
$y = 57 \frac{1}{4}$

Vertex is $\left(- \frac{17}{2} , 57 \frac{1}{4}\right)$

The vertex form of the quadratic equation is -

$y = a {\left(x - h\right)}^{2} + k$

Where -

$a = - 1$ Coefficient of ${x}^{2}$
$h = - \frac{17}{4}$ $x$ co-ordinate of the vertex
$k = 57 \frac{1}{4}$ $y$ co-ordinate of the vertex

Now substitute these values in the vertex formula.

$y = - 1 {\left(x - \left(- \frac{17}{4}\right)\right)}^{2} + \left(57 \frac{1}{4}\right)$

$y = - 1 {\left(x + \frac{17}{4}\right)}^{2} + 57 \frac{1}{4}$