91影视

$$\begin{gathered} f\left(x\right)=a{x}^2+bx鈭�c \\ f\left(x\right)=3x^2+12x鈭�12 \end{gathered}$$

Since a is 3 , the parabola opens upward.

$$\begin{gathered} x=-\frac{b}{2a} \\ x=-\frac{12}{2\left(3\right)} \\ x=-2 \end{gathered}$$

The axis of symmetry is $x=-2$.

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

Find$f(-2)$

role="math" localid="1645169868337" $$\begin{gathered} f\left(x\right)=3x^2+12x鈭�12 \\ f(-2)=3{(-2)}^2+12(-2)鈭�12 \\ f(-2)=12-24鈭�12 \\ f(-2)=-24 \end{gathered}$$

The vertex is $(-2,-24)$.

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

$$\begin{gathered} f\left(x\right)=3x^2+12x鈭�12 \\ f\left(0\right)=3{\left(0\right)}^2+12\left(0\right)鈭�12 \\ f\left(0\right)=鈭�12 \end{gathered}$$

The y-intercept is (0, 鈭�12).

The point (0, 鈭�12) is two units to the right of the line of symmetry.

The point two units to the left of the line of symmetry is $(-4,鈭�12)$.

Point symmetric to the y-intercept is $(-4,鈭�12)$

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

$$\begin{gathered} f\left(x\right)=3x^2+12x鈭�12 \\ 0=3x^2+12x鈭�12 \end{gathered}$$

Factor the GCF:

role="math" localid="1645171288550" $$\begin{gathered} 0=3(x^2+4x鈭�4) \\ 0=x^2+4x鈭�4 \end{gathered}$$

Solve the trinomial:

role="math" localid="1645171453723" $x=-4.828andx=0.828$

The x-intercepts are $(-4.828,0)and(0.828,0)$

We graph the vertex, intercepts, and the point symmetric to the y-intercept.

We connect these 5 points to sketch the parabola.