first off, is good to notice that on f(x) and g(x) the squared variable is the x, therefore the parabolas are vertical, and h(x) is well, we can see is vertical as well, so their axis of symmetry will come from the x-coordinate of their vertex.
f(x) is in vertex form already, so its vertex is clearly
[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\qquadvertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\-------------------------------\\\\f(x)=-2(x-\stackrel{h}{4})^2+\stackrel{k}{2}\qquad vertex~(\stackrel{x}{4},2)\qquad \qquad \stackrel{\textit{axis of symmetry}}{x=4} [/tex]
now for g(x),
[tex] \bf \textit{vertex of a vertical parabola, using coefficients}\\\\g(x)=\stackrel{\stackrel{a}{\downarrow }}{5}x^2\stackrel{\stackrel{b}{\downarrow }}{-10}x\stackrel{\stackrel{c}{\downarrow }}{+7}\qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)\\\\\\\left(-\cfrac{-10}{2(5)}~~,~~7-\cfrac{(-10)^2}{4(5)} \right)\implies (\stackrel{x}{1}~,2)\qquad \qquad \stackrel{\textit{axis of symmetry}}{x=1} [/tex]
and for h(x) its vertex or U-turn is at -2,2, so the axis of symmetry for that will be x = -2.
so, from smallest to largest... well , -2, 1, 4, namely h(x), g(x), f(x).
