Quadratic function problem - maximum height - frogs leap

SOLUTION ==========================
$h(x)=\displaystyle -{1\over 3}x^2+{4\over 3}x+4$ 

(a) We need to know where the maximum occurs, this will be on the axis of symmetry. 
You may know the formula $$x={-b\over 2a}$$ for the axis of symmetry, which gives 
$$x={-(4/3)\over 2(-1/3)}=2$$  

[you could also see this by factorising $$h(x)=-1/3(x^2-4x-12)=-1/3(x+2)(x-6)$$ and so the max occurs half way between the x-intercepts which are $x=-2, x=6$, so max occurs at $x =(-2+6)/2=2$ as before.] 

So the max occurs when $x=2$, and since the frog starts at $x=0$, then it reached the maximum height after 2 ft. 

(b) When $x=2$,  
$h(2)=-1/3 (2)^2+4/3(2)+4$
$= -4/3+8/3+4$
$=5 1/3$ or ${16\over 3}$ ft