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$y=−x^2+10x−13$with the standard quadratic function $y=a{x}^2+bx+c$.

$a=−1,\text{ }b=10,\text{ }c=−13$

Substitute$a=−1$and$b=10$in $x=−\frac{b}{2a}$.

$$\begin{gathered} x=−\left(\frac{10}{2\left(−1\right)}\right) \\ x=−\left(\frac{10}{−2}\right) \\ x=−\left(−5\right) \\ x=5 \end{gathered}$$

So, the axis of symmetry is given by

$x=5$

Since, $a>0$.

So, the function has a minimum value at $x=5$.

Substitute$x=5$in $y=−x^2+10x−13$.

$$\begin{gathered} y=−{\left(5\right)}^2+10\left(5\right)−13 \\ y=−25+50−13 \\ y=50−38 \\ y=12 \end{gathered}$$

So, the vertex point is located at the point $\left(5,\text{ }12\right)$.

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

So, the $y$-intercept is localid="1647852521909" $-13$.

Therefore the vertex is $\left(5,\text{ }12\right)$, the equation of the axis of symmetry is$x=5$and the $y$-intercept is $−13$.