let \(y=x^{2}-5 x+6\)y completing the square method, we add and subtract the square of half the coeffi. cient of \(x\) to the right hand side of the equation
i.e. \(\left(\frac{-5}{2}\right)^{2}=\frac{25}{4}\)
\begin{aligned}
\Rightarrow y &=x^{2}-5 x-\frac{25}{4}+\frac{25}{4}+6 \\
y &=x^{2}-5 x+\frac{25}{4}-\frac{25}{4}+6 \\
y &=\left(x-\frac{5}{2}\right)^{2}+\left(-\frac{25}{4}+6\right) \\
y &=\left(x-\frac{5}{2}\right)^{2}+\left(-\frac{1}{4}\right) \\
\Rightarrow x &=\frac{5}{2}
\end{aligned}
alternatively, by differentiation,
\begin{array}{l}
y=x^{2}-5 x+6 \\
\frac{d y}{d x}=2 x-5
\end{array}
at turning point:minimum or maximum point:
\begin{array}{l}
\frac{d y}{d x}=0 \\
\Rightarrow 2 x-5=0, x=\frac{5}{2}
\end{array}

for minimum point:
\(\frac{d^{2} y}{d x^{2}}>0\) and for maximum point,
\(\frac{d^{2} y}{d x^{2}}<0\)
from \(\frac{d y}{d x}=2 x-5\), differentiating further
\(\frac{d^{2} y}{d x^{2}}=2>0\) which is gleater than 0
this implies that a mmimum point occur at \(x=\frac{5}{2}\)