based on that table of values, we know the highest point(vertex) is at 0.9, 5.9
we can also use just one another point... say 1.4, 4.6
[tex]\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
y=a(x-{{ h}})^2+{{ k}}
\end{array} \qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\[/tex]
[tex]\bf \begin{cases}
h=0.9\\
k=5.9
\end{cases}\implies y=a(x-0.9)^2+5.9
\\\\\\
\begin{cases}
\textit{another point}\\
x=1.4\\
y=4.6
\end{cases}\implies 4.6=a(1.4-0.9)^2+5.9
\\\\\\
4.6=a(0.5)^2+5.9\implies 4.6-5.9=a(0.5)^2\implies -1.3=a(0.5)^2
\\\\\\
\cfrac{-1.3}{0.5^2}=a\implies -5.2=a
\\\\\\
thus\qquad \qquad \boxed{y=-5.2(x-0.9)^2+5.9}[/tex]
