The given graph is given as :
Since , I have to find the equation for $g(x)$ , and since $g(x)$ is parabolic , I figured that the equation used will be Let $g(x) = y$, then the equation becomes $(y-k)^2 = 4a(x-h)$ , which further on solving for y becomes $y=\sqrt{4a(x-h)}+k$ , hence we have three unknowns $a,h$ and $k$.
I can't figure out how to proceed from here , also what help (if any) does $f(x)$ provides me with?
From the graph, $g(x)$ passes through the points $(6,6) (0,2) (-2, 0)$. Insert these points into the equation $(y-k)^2=4a(x-h)$ : $$(6-k)^2=4a(6-h)$$ $$(2-k)^2=4a(0-h)$$ $$(0-k)^2=4a(-2-h)$$ Subtract the 2nd and 3rd equations from the 1st, using the identity $a^2-b^2=(a-b)(a+b)$ on the LHS : $$(4)(8-2k)=4a(6) \implies 4-k=3a$$ $$(6)(6-2k)=4a(8) \implies 9-3k=8a$$ Multiply the 4th eqn by 3 and subtract the 5th eqn to find $a$, then back-substitute to find $k, h$ : $$3=a$$ $$k=4-3a=4-9=-5$$ $$-4ah=(2-k)^2 \implies -12h=49 \implies h=-\frac{49}{12}$$