91影视

First, separate the xterm from the constant and factor, we get:

$$\begin{gathered} f\left(x\right)=3x^2-6x+5 \\ f\left(x\right)=3\left(x^2-2x\right)+5 \end{gathered}$$

Now, Take half of $2$and then square it, we get:

${\left(\frac{1}{2}路2\right)}^2=1$

So, $1$will complete the square.

The constant $1$completes the square and then we have to subtract $3脳1=3$from the constant.

role="math" $$\begin{gathered} f\left(x\right)=3\left(x^2-2x\right)+5 \\ f\left(x\right)=3\left(x^2-2x+1\right)+5-3 \\ f\left(x\right)=3{\left(x-1\right)}^2+2\_\_\_\_\left(1\right)\left[{\left(a-b\right)}^2=a^2-2ab+b^2\right] \end{gathered}$$

From the equation (1), we get:

$$\begin{gathered} a=3 \\ h=1 \\ k=2 \end{gathered}$$

Since the value of$a$is positive then it is an upwards parabola

The vertex of the parabola from the equation is given as:

$\left(h,k\right)=\left(1,2\right)$

Substituting $x=0$in the equation (1), we get:

$$\begin{gathered} f\left(x\right)=3{\left(x-1\right)}^2+2 \\ f\left(0\right)=3{\left(0-1\right)}^2+2 \\ f\left(0\right)=3+2 \\ f\left(0\right)=5 \end{gathered}$$

The y- intercept is$\left(0,5\right)$

the point symmetric to y-intercept across the axis of symmetry is$\left(2,5\right)$

The discriminant is negative, so there are

no x-intercepts. Graph the parabola

Using all the properties of the function, the graph is given as: