91Ó°ÊÓ

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

$a=5,\text{ }b=20,\text{ }c=10$

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 $\left(−2,\text{ }−10\right)$.

Since, the y-intercept is given by c.

So, the y-intercept is 10.

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