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

The graph of the function $y=a{x}^2+bx+c,$

Opens upward and has a minimum value at $x=−\frac{b}{2a}$, when $a>0$.

Opens downward and has a maximum value at $x=−\frac{b}{2a}$, when $a<0$.

The maximum or minimum point of the function $y=a{x}^2+bx+c$,is called the vertex.

Compare the quadratic function $y=x^2−5x+4$with the standard equation of the quadratic function, $y=a{x}^2+bx+c$.

$a=1,\text{ }b=−5,\text{ }c=4$

Substitute, $a=1$and role="math" localid="1647752801168" $b=-5$in $x=−\frac{b}{2a}$.

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

Since,

So, the graph of the function opens upward and has a minimum point at $x=\frac{5}{2}$.

Substitute $x=\frac{5}{2}$in $y=x^2−5x+4$.

$y={\left(\frac{5}{2}\right)}^2−5\left(\frac{5}{2}\right)+4$

$$\begin{gathered} =\frac{25}{4}−\frac{25}{2}+4 \\ =\frac{25−2\left(25\right)+4\left(4\right)}{4} \\ =\frac{25−50+16}{4} \\ =\frac{41−50}{4} \\ =−\frac{9}{4} \end{gathered}$$

Hence, vertex $=\left(\frac{5}{2},\text{ }−\frac{9}{4}\right)$

Therefore, the coordinate of the vertex is $\left(\frac{5}{2},\text{ }−\frac{9}{4}\right)$.