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

$y=5x^2+20x+10$

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$y=5x^2+20x+10$ with the standard quadratic function $y=a{x}^2+bx+c$.

$a=5,b=20,c=10$

Substitute$a=5$ and$b=20$ in $x=-\frac{b}{2a}$.

$$\begin{gathered} x=-\left(\frac{20}{2\left(5\right)}\right) \\ x=-\left(\frac{20}{10}\right) \\ x=-\left(2\right) \\ x=-2 \end{gathered}$$

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 $y=5x^2+20x+10$.

$$\begin{gathered} y=5{\left(-2\right)}^2+20\left(-2\right)+10 \\ y=5\left(4\right)-40+10 \\ y=20-40+10 \\ y=30-40 \\ y=-10 \end{gathered}$$

So, the vertex point is located at the point $\left(-2,-10\right)$.

Since, the y-intercept is given by c.

So, the y-intercept is 10.

Therefore the vertex is $\left(-2,-10\right)$, the equation of the axis of symmetry is$x=-2$ and the y-intercept is 10.