91Ó°ÊÓ

The inequality in standard form is$x^2+2x-8<0$,

For a quadratic function of the form$a{x}^2+bx+c$if $a>0$, then it represents a parabola open upwards.

Compare the function with

Here, $a=1,b=2,andc=-8$

So, the function represents a parabola open upward,

The axis of symmetry of a parabola$a{x}^2+bx+c$is given by $x=-\frac{b}{2a}$.

Now, evaluate the axis of symmetry of the parabola.

$$\begin{gathered} x=-\frac{2}{2\left(1\right)} \\ x=-1 \end{gathered}$$

The axis of symmetry is the line $x=-1$.

The vertex is given by $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$

Calculate the vertex of the parabola as shown:

$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^2+2\left(-1\right)-8 \\ =1-2-8 \\ =-9 \end{gathered}$$

The vertex is $\left(-1,-9\right)$

Now, for the y-intercept of the parabola find $f\left(0\right)$.

So, the $y$intercept of the parabola is

$$\begin{gathered} f\left(0\right)={\left(0\right)}^2+2\left(0\right)-8 \\ f\left(0\right)=-8 \end{gathered}$$

Now, for x-intercept solve$f\left(x\right)=0$

Use factorization of quadratic equation.

$$\begin{gathered} x^2+2x-8=0 \\ {x}^2+4x-2x-8=0 \\ \left(x+4\right)\left(x-2\right)=0 \\ x=-4,x=2 \end{gathered}$$

So, the x-intercepts are $(-4,0)and(2,0)$.

Now, graph the quadratic function by using the vertex, axis of symmetry, $x$intercept, and $y$- intercepts.

Now, determine the solution from the graph.

The inequality asks for the values of $x$for which the function is less than 0.

$-4<x<2$

The solution in interval notation is $\left(-4,2\right)$.

The values $-4$and 2 are not included since the inequality is less than type.