Question 1105122
If the graph passes through (4,0), then
{{{0=a(4-h)^2}}} , so either {{{h=4}}} or {{{a=0}}} .
However, if that function is really a quadratic it must be true that
{{{a<>0}}} .
Otherwise, the function would just be {{{y=0}}} .
So , {{{highlight(h=4)}}} .
Substituting that value, and the coordinates of (8,-4), we get
{{{-4=a(8-4)^2}}}
{{{-4=a*4^2}}}
{{{-4=16a}}}
{{{a=16/(-4)}}}
{{{highlight(a=-1/4)}}}
Three quadratic function is
{{{highlight(y=(-1/4)(x-4)^2)}}} or {{{highlight(y=-(x-4)^2/4)}}} .