# How do you find the vertex and the intercepts for f(x)= -x^2-6x-5?

May 4, 2017

see explanation.

#### Explanation:

$\text{for the standard quadratic function } y = a {x}^{2} + b x + c$

"the x-coordinate of the vertex " x_(color(red)"vertex")=-b/(2a)

$\text{for the function } y = - {x}^{2} - 6 x - 5$

$\text{then " a=-1,b=-6" and } c = - 5$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 6}{- 2} = - 3$

$\text{substitute this value into the function and evaluate for y}$

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = - {\left(- 3\right)}^{2} - 6 \left(- 3\right) - 5 = 4$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- 3 , 4\right)$

$\textcolor{b l u e}{\text{to find intercepts}}$

• " let x=0, in function, for y-intercept"

• " let y=0, in function, for x-intercepts"

$x = 0 \to y = - 5 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to - {x}^{2} - 6 x - 5 = 0$

$\Rightarrow {x}^{2} + 6 x + 5 = 0 \leftarrow \text{ multiply through by - 1}$

$\left(x + 5\right) \left(x + 1\right) = 0$

$\Rightarrow x = - 5 , x = - 1 \leftarrow \textcolor{red}{\text{ x-intercepts}}$
graph{-x^2-6x-5 [-10, 10, -5, 5]}