Question 812727
Parabola for simplicity, {{{y=a(x-h)^2+k}}} standard form.


Vertex means, {{{y=a(x-1)^2+5}}}
If you make no assumptions about "a", try the given two intercepts in the equation:


0=a(-5-1)^2+5
-6a+5=0
-6a=-5
a=5/6


{{{a(7-1)^2+5=0}}}
{{{6a+5=0}}}
{{{6a=-5}}}
{{{a=-5/6}}}---------we will want THIS, because:


....Thinking carefully about how the graph should appear, vertex (1,5) must be a maximum because it is above the two x-intercepts, so {{{a<0}}} and we must choose {{{a=-5/6}}}


Our equation, parabola in standard form be {{{highlight(y=-(5/6)(x-1)^2+5)}}}