Consider $y=x^2-5x+4$

Find the coordinates of the vertex. Is it a maximum or minimum point?

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

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, $y=a{x}^2+bx+c$.

$a=1,b=-5,c=4$

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 $=\left(\frac{5}{2},-\frac{9}{4}\right)$

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