91Ó°ÊÓ

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(x\right)=4x^2+24x+36 \end{gathered}$$

Since a is 4 , the parabola opens upward.

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{24}{2\left(4\right)} \\ x=-3 \end{gathered}$$

The axis of symmetry is$x=-3$

The vertex is on the line$x=-3$

Find $f(-3)$

$$\begin{gathered} f\left(x\right)=4x^2+24x+36 \\ f(-3)=4{(-3)}^2+24(-3)+36 \\ f(-3)=36-72+36 \\ f(-3)=0 \end{gathered}$$

The vertex is$(-3,0)$

The y-intercept occurs when x = 0 . Find f (0) .

$$\begin{gathered} f\left(x\right)=4x^2+24x+36 \\ f\left(0\right)=4{\left(0\right)}^2+24\left(0\right)+36 \\ f\left(0\right)=36 \end{gathered}$$

The y-intercept is $(0,36)$

The point (0, 36) is three units to the right of the line of symmetry.

The point three units to the left of the line of symmetry is $(-6,36)$

Point symmetric to the y-intercept is$(-6,36)$

The x-intercept occurs when f (x) = 0

$$\begin{gathered} f\left(x\right)=4x^2+24x+36 \\ 0=4x^2+24x+36 \end{gathered}$$

Factor the GDF:

localid="1645174243068" $$\begin{gathered} 0=4(x^2+6x+9) \\ 0=x^2+6x+9 \end{gathered}$$

Factor the trinomial:

localid="1645174304788" $$\begin{gathered} 0={(x+3)}^2 \\ -3=x \end{gathered}$$

The x-intercept is$(-3,0)$

Connect the points to graph the parabola.