# How do you find the axis of symmetry and vertex point of the function: y = -x^2 - x + 9?

Jun 10, 2018

Vertex: $\left(- \frac{1}{2} , 9 \frac{1}{4}\right)$
Line of symmetry: $x = - \frac{1}{2}$

#### Explanation:

use $- \frac{b}{2 a}$ to find the vertex:

$- \frac{- 1}{2 \times \left(- 1\right)} = - \frac{1}{2}$

plug $- \frac{1}{2}$ in for $x$

$= 9 \frac{1}{4}$

vertex: $\left(- \frac{1}{2} , 9 \frac{1}{4}\right)$

on parabolic functions the axis of symmetry is the vertical line where the vertex is, so the equation is:

$x = - \frac{1}{2}$

An alternate way, using the vertex form:

#### Explanation:

We can change the standard form of the equation to the vertex form, which has the general form:

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

We get there by completing the square (we do that by taking the constant of the $x$ term, halving it, squaring the half, then adding and subtracting it into the equation:

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

$y = - \left({x}^{2} + x\right) + 9$

$y = - \left({x}^{2} + x + \frac{1}{4} - \frac{1}{4}\right) + 9$

$y = - \left({\left(x + \frac{1}{2}\right)}^{2} - \frac{1}{4}\right) + 9$

$y = - {\left(x + \frac{1}{2}\right)}^{2} + \frac{1}{4} + 9$

$y = - {\left(x + \frac{1}{2}\right)}^{2} + 9 \frac{1}{4}$

The vertex is given by $\left(h , k\right)$, which in this case is $\left(- \frac{1}{2} , 9 \frac{1}{4}\right)$.

The axis of symmetry runs through the vertex vertically and so takes the form of $x = \text{the " x " value of the vertex}$, which in this case is

$x = - \frac{1}{2}$

This can all be seen in the graph:

graph{-x^2-x+9[-3,3,8,10]}

note that I've adjusted the graph unequally vertically and horizontally to better show the vertex