so let's use the vertex form of a parabola, since we know what its vertex coordinates are, let's use those, and then expand from there, so from the back to the front.
[tex] \bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
\boxed{y=a(x- h)^2+ k}\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{\frac{5}{2}}{ h},\stackrel{-\frac{1}{4}}{ k})\\\\
------------------------------- [/tex]
[tex] \bf y=1\left(x-\cfrac{5}{2} \right)^2-\cfrac{1}{4}\implies y=1\stackrel{FOIL}{\left(x-\cfrac{5}{2} \right)\left(x-\cfrac{5}{2} \right)}-\cfrac{1}{4}
\\\\\\
y=\stackrel{FOIL}{\left(x^2-5x+\cfrac{5^2}{2^2}\right)}-\cfrac{1}{4}\implies y=x^2-5x+\cfrac{25}{4}-\cfrac{1}{4}
\\\\\\
y=x^2-5x+\cfrac{24}{4}\implies y=\stackrel{\stackrel{a}{\downarrow }}{1}x^2\stackrel{\stackrel{b}{\downarrow }}{-5}x+6 [/tex]
