91Ó°ÊÓ

Consider the given function,

$f\left(x\right)=3x^2-12x+7$

Rewrite the function in $f\left(x\right)=a{\left(x-h\right)}^2+k$form by completing the square,

localid="1645120743155" $$\begin{gathered} f\left(x\right)=3\left(x^2-4x\right)+7 \\ f\left(x\right)=3\left(x^2-4x+4\right)+7-12 \\ f\left(x\right)=3{\left(x-2\right)}^2-5......\left(i\right) \end{gathered}$$

Identify the constants,

$$\begin{gathered} a=3, \\ h=2, \\ k=-5 \end{gathered}$$

Since $a=3$, the parabola opens upward.

The axis of symmetry is $x=h$. Then,

localid="1645120777400" $x=2$.

The vertex is $\left(h,k\right)$. Then localid="1645120785853" $\left(2,-5\right)$.

Substitute $x=0$in equation (i),

localid="1645120811830" $$\begin{gathered} f\left(0\right)=3{\left(0-2\right)}^2-5 \\ =3\left(4\right)-5 \\ =7 \end{gathered}$$

Therefore, y-intercept is localid="1645120818466" $\left(0,7\right)$.

The point symmetric to $\left(0,7\right)$across the axis of symmetry is $\left(4,7\right)$.

Substitute $f\left(x\right)=0$in equation (i),

localid="1645205556999" $$\begin{gathered} 0=3{\left(x-2\right)}^2-5 \\ 0=3x^2-12x+7 \\ x=\frac{12+\sqrt{60}}{6},\frac{12-\sqrt{60}}{6} \\ x≈0.7,3.3 \end{gathered}$$

Therefore, the x-intercepts are $\left(0.7,0\right),\left(3.3,0\right)$.

Plot the graph using the obtained information,