# How do you solve quadratic inequality, graph, and write in interval notation  -x^2 - x + 6 <0?

Aug 3, 2017

$\left(- \infty , - 3\right) \cup \left(2 , + \infty\right)$

#### Explanation:

$\text{graph the parabola } y = - {x}^{2} - x + 6$
$\text{and consider which parts are less than zero, that is below}$
$\text{the x-axis}$

$\textcolor{b l u e}{\text{finding the x and y intercepts}}$

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

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

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

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

$\text{multiply through by - 1}$

$\Rightarrow {x}^{2} + x - 6 = 0$

$\text{the factors which multiply to give - 6 and sum to + 1}$
$\text{are + 3 and - 2}$

$\Rightarrow \left(x + 3\right) \left(x - 2\right) = 0$

$\Rightarrow x = - 3 \text{ or "x=2larrcolor(red)" x-intercepts}$

$\textcolor{b l u e}{\text{obtaining the shape of the parabola}}$

• " if "a>0" then minimum " uuu

• " if "a<0" then maximum " nnn

$\text{for } y = - {x}^{2} - x + 6 \textcolor{w h i t e}{x} a < 0$

$\text{we can now graph the parabola}$
graph{-x^2-x+6 [-10, 10, -5, 5]}

$x < - 3 \text{ or " x>2 " are the parts below the x-axis}$

$\text{ in interval notation} \left(- \infty , - 3\right) \cup \left(2 , + \infty\right)$