# How do you find the vertex and the intercepts for f(x)= -x^2-3x-6 ?

Oct 4, 2017

$\text{see explanation}$

#### Explanation:

"given the equation of a parabola in standard form "

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)

$f \left(x\right) = - {x}^{2} - 3 x - 6 \text{ is in standard form}$

$\text{with } a = - 1 , b = - 3 , c = - 6$

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

$\text{substitute this value into f(x) for y-coordinate}$

$\Rightarrow {y}_{\textcolor{red}{\text{vertex}}} = - {\left(- \frac{3}{2}\right)}^{2} - 3 \left(- \frac{3}{2}\right) - 6 = - \frac{15}{4}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex}} = \left(- \frac{3}{2} , - \frac{15}{4}\right)$

$\textcolor{b l u e}{\text{Intercepts}}$

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

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

$x = 0 \to y = - 6 \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to - {x}^{2} - 3 x - 6 = 0$

$\text{checking the value of the "color(blue)"discriminant}$

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4 \times - 1 \times - 6\right) = - 15$

$\text{since "Delta<0" then no real solutions}$

$\Rightarrow f \left(x\right) \text{ does not intersect with the x-axis}$
graph{-x^2-3x-6 [-10, 10, -5, 5]}