Graph the function.

$y=3x^2-12x+5$

A quadratic function, which is written in the form, $y=a{x}^2+bx+c$, where,$a\ne 0$ is called the standard form of the quadratic function.

For the function$y=a{x}^2+bx+c$,

(1) If $a>0$, then the function has the minimum value at$x=-\frac{b}{2a}$ and the vertex is located at the minimum point.

(2) If $a<0$, then the function has the maximum value at$x=-\frac{b}{2a}$ and the vertex is located at the maximum point.

Opens upward and has a minimum value at $x=-\frac{b}{2a}$, when.

Opens downward and has a maximum value at $x=-\frac{b}{2a}$, when.

The $y$-intercept of the function $y=a{x}^2+bx+c$is always at $c$

Compare the quadratic function $y=3x^2-12x+5$with the standard quadratic function $y=a{x}^2+bx+c$.

$a=3,b=-12,c=5$

Substitute$a=3$and$b=-12$in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{-12}{2\left(3\right)}\right) \\ x=-\left(\frac{-12}{6}\right) \\ x=-\left(-2\right) \\ x=2 \end{gathered}$$

Since, $a>0$.

So, the graph opens upward and has a minimum value at $x=2$.

Since, the $y$-intercept is given by $c$.

So, the $y$-intercept is 5.

The graph of the function $y=3x^2-12x+5$is shown below.