# How do you find the VERTEX of a parabola f(x)= -x^2-6x-5?

Jul 30, 2015

Convert the given equation into vertex form [$y = m {\left(x - a\right)}^{2} + b$]
to get $\left(x , f \left(x\right)\right) = \left(- 3 , 4\right)$
or use the formula (see below) for the $f \left(x\right)$ vertex coordinate and solve for $x$

#### Explanation:

Vertex form
An equation in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = m {\left(x - a\right)}^{2} + b$
has its vertex at $\left(a , b\right)$

Given: $f \left(x\right) = - {x}^{2} - 6 x - 5$

Extract $m$ for the vertex form:
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- 1\right) \left({x}^{2} + 6 x\right) - 5$

Complete the square
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- 1\right) \left({x}^{2} + 6 x + 9\right) - 5 - \left(- 1\right) \left(9\right)$
Simplifying
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = \left(- 1\right) {\left(x + 3\right)}^{2} + 4$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$which is in vertex form with the vertex at $\left(- 3 , 4\right)$

y vertex coordinate formula
An equation in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = a {x}^{2} + b x + c$
has the $f \left(x\right)$ coordinate at
$\textcolor{w h i t e}{\text{XXXX}}$$c - {b}^{2} / \left(4 a\right)$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$[this is a result of the significance of the discriminant]

For $f \left(x\right) = - {x}^{2} - 6 x - 5$
$\textcolor{w h i t e}{\text{XXXX}}$$a = - 1$$\textcolor{w h i t e}{\text{XXXX}}$$b = - 6$$\textcolor{w h i t e}{\text{XXXX}}$$c = - 5$
and the $f \left(x\right)$ coordinate of the vertex is at
$\textcolor{w h i t e}{\text{XXXX}}$$- 5 - \left({\left(- 6\right)}^{2} / \left(4 \left(- 1\right)\right)\right) = 4$

Solving for the corresponding $x$ coordinate:
$\textcolor{w h i t e}{\text{XXXX}}$$- {x}^{2} - 6 x - 5 = 4$

$\textcolor{w h i t e}{\text{XXXX}}$$- {x}^{2} - 6 x - 9 = 0$

$\textcolor{w h i t e}{\text{XXXX}}$$\left({x}^{2} + 6 x + 9\right) = 0$

$\textcolor{w h i t e}{\text{XXXX}}$${\left(x + 3\right)}^{2} = 0$

$\textcolor{w h i t e}{\text{XXXX}}$$\rightarrow$$\textcolor{w h i t e}{\text{XXXX}}$$x = - 3$

and again, we have the vertex is at $\left(- 3 , 4\right)$