now, notice g(x), and recall that
[tex]\bf \qquad \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\ ({{ h}},{{ k}})[/tex]
that simply means that the vertex for g(x) is at [tex]\bf g(x)=4(x-\stackrel{h}{8})^2+\stackrel{k}{9}\impliedby (8,9)[/tex]
both parabolas are opening upwards, so they look like a "bowl", from up it goes down down down and reaches the vertex the goes back up up up.
the lowest point or vertex is a minimum, as low as it gets, so, which one is greater then? f(x)'s which is (9, 8) or g(x)'s which is (8, 9) ?
