[tex]\text{Let}\ f(x)=ax^2+bx+c,\ \text{then}\ f(x)=a(x-p)^2+k,\ \text{where}\\\\h=\dfrac{-b}{2a},\ k=f(h)\\\\-----------------\\\\\text{We have}\ f(x)=2x^2-5x+1\to a=2,\ b=-5,\ c=1.\\\\\text{Substitute:}\\\\h=\dfrac{-(-5)}{2(2)}=\dfrac{5}{4}\\\\k=f\left(\dfrac{5}{4}\right)=2\left(\dfrac{5}{4}\right)^2-5\left(\dfrac{5}{4}\right)+1=2\left(\dfrac{25}{16}\right)-\dfrac{25}{4}+1\\\\=\dfrac{25}{8}-\dfrac{25\cdot2}{4\cdot2}+\dfrac{8}{8}=\dfrac{25-50+8}{8}=-\dfrac{17}{8}\\\\\text{Therefore we have}[/tex]
[tex]f(x)=2\left(x-\dfrac{5}{4}\right)-\dfrac{17}{8}=\boxed{2\left(x+\left(-\dfrac{5}{4}\right)\right)^2+\left(-\dfrac{17}{8}\right)}[/tex]
[tex]q=-\dfrac{5}{4},\ k=-\dfrac{17}{8}[/tex]
