Question: Find the vertex, the equation of the axis of symmetry, and the $y$-intercept of the given equation.

$y=2x^2-12x+6$

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.

The axis of symmetry for the function$y=a{x}^2+bx+c$ is

$x=-\frac{b}{2a}$

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

Compare the quadratic function$y=2x^2-12x+6$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=2,b=-12,c=6$

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

$$\begin{gathered} x=-\left(\frac{-12}{2\left(2\right)}\right) \\ x=-\left(\frac{-12}{4}\right) \\ x=-\left(-3\right) \\ x=3 \end{gathered}$$

So, the axis of symmetry is given by

$x=3$

Since, $a>0$.

So, the function has a minimum value at $x=3$.

Substitute$x=3$ in $y=2x^2-12x+6$.

$$\begin{gathered} y=7{\left(3\right)}^2-12\left(3\right)+6 \\ y=7\left(9\right)-36+6 \\ y=63-36+6 \\ y=69-36 \\ y=33 \end{gathered}$$

So, the vertex point is located at the point $\left(3,33\right)$.

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

So, the $y$-intercept is 6.

Therefore the vertex is $\left(3,33\right)$, the equation of the axis of symmetry is$x=3$ and the $y$-intercept is 6.