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

$y=7x^2-28x+14$

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$y=7x^2-28x+14$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=7,b=-28,c=14$

Substitute$a=7$ and$b=-28$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{-28}{2\left(7\right)}\right) \\ x=-\left(\frac{-28}{14}\right) \\ x=-\left(-2\right) \\ x=2 \end{gathered}$$

So, the axis of symmetry is given by

$x=2$

Since, $a>0$.

So, the function has a minimum value at $x=2$.

Substitute$x=2$ in $y=7x^2-28x+14$.

$$\begin{gathered} y=7{\left(2\right)}^2-28\left(2\right)+14 \\ y=7\left(4\right)-56+14 \\ y=28-56+14 \\ y=42-56 \\ y=-14 \end{gathered}$$

So, the vertex point is located at the point $\left(2,-14\right)$.

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

So, the $y$-intercept is 14.

Therefore the vertex is $\left(2,-14\right)$, the equation of the axis of symmetry is$x=2$ and the $y$-intercept is 14.