91Ó°ÊÓ

Look at a in equation $f\left(x\right)=x^2−8x+12$

Since a is positive, the parabola opens upward.

$$\begin{gathered} f\left(x\right)=x^2−8x+12 \\ a=1,b=-8,c=-8 \end{gathered}$$

The parabola opens upwards.

$f\left(x\right)=x^2−8x+12$

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

Axis of symmetry is

$$\begin{gathered} x=-\frac{-8}{2\left(1\right)} \\ x=4 \end{gathered}$$

The axis of symmetry is the line$x=4$

The vertex is on the axis of symmetry.

Substitute $x=4$into a function.

Vertex:

$$\begin{gathered} f\left(x\right)=x^2−8x+12 \\ f\left(4\right)={\left(4\right)}^2−8\left(4\right)+12 \\ f\left(4\right)=16−32+12 \\ f\left(4\right)=-4 \end{gathered}$$

The vertex is$(4,-4)$

We Find f(0)

We use the axis of symmetry to find a point symmetric to the y-intercept. The y- intercept is 4 units left to the axis of symmetry, $x=4$

A point 4 units to the right of the axis of symmetry has $x=8$

y-intercept:

$$\begin{gathered} f\left(x\right)=x^2−8x+12 \\ f\left(0\right)={\left(0\right)}^2−8\left(0\right)+12 \\ f\left(0\right)=12 \end{gathered}$$

The y-intercept is $(0,12)$

Point symmetric to y-intercept:

The point is$(8,12)$

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