91Ó°ÊÓ

Look at a in equation $f\left(x\right)=x^2+2x−8$

Since a is positive, the parabola opens upward.

$f\left(x\right)=x^2+2x−8$

$a=1,b=2,c=-8$

The parabola opens upwards.

$f\left(x\right)=x^2+2x−8$

The axis of symmetry is the line $x=-\frac{b}{2a}$.

Axis of symmetry is

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

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

The vertex is on the axis of symmetry.

Substitute $x=-1$into the function.

Vertex:

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

The vertex is$(-1,-9)$

We Find f(0)

We use the axis of symmetry to find a point symmetric to the y-intercept. The y- intercept is 1 unit right to the axis of symmetry, $x=-1$.

A point 1 unit to the left of the axis of symmetry has$x=-2$

y-intercept:

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

The y-intercept is $(0,-8)$

Point symmetric to y-intercept:

The point is$(-2,-8)$

We solve $f\left(x\right)=0$

We can solve this quadratic equation by factoring.

x-intercepts:

role="math" localid="1645164291991" $$\begin{gathered} f\left(x\right)=x^2+2x−8 \\ 0=x^2+2x−8 \\ 0=(x+4)(x-2) \\ 0=x+4;0=x-2 \\ -4=x;2=x \end{gathered}$$

The x-intercepts are$(-4,0)and(2,0)$

We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola.