91Ó°ÊÓ

$$\begin{gathered} f\left(x\right)=a{x}^2+bx+c \\ f\left(x\right)=−3x^2−6x−4 \end{gathered}$$

Since a is -3 , the parabola opens downward.

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{-6}{-3} \\ x=-2 \end{gathered}$$

The axis of symmetry is $x=-2$

The vertex is on the line $x=-2$

Find $f(-2)$

$$\begin{gathered} f\left(x\right)=−3x^2−6x−4 \\ f(-2)=−3{(-2)}^2−6(-2)−4 \\ f(-2)=−12+12−4 \\ f(-2)=−4 \end{gathered}$$

The vertex is$(-2,4)$

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

$$\begin{gathered} f\left(x\right)=−3x^2−6x−4 \\ f\left(0\right)=−3{\left(0\right)}^2−6\left(0\right)−4 \\ f\left(0\right)=−4 \end{gathered}$$

The y-intercept is $(0,-4)$

The point $(0,-4)$is one unit to the right of the line of symmetry.

The point one unit to the left of the line of symmetry is $(-2,-4)$

Point symmetric to the y-intercept is$(-2,-4)$

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

Find f (x) = 0 .

$$\begin{gathered} f\left(x\right)=−3x^2−6x−4 \\ 0=−3x^2−6x−4 \end{gathered}$$

Test the discriminant.

$$\begin{gathered} b^2-4ac \\ {(-6)}^2-4(-3)(-4) \\ 36-48 \\ -12 \end{gathered}$$

Since the value of the discriminant is negative, there is no real solution and so no x-intercept.

Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.