Question 956636
(0,0)
{{{0=0+0+c}}}
{{{highlight(c=0)}}}
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.
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(7/2,4)
{{{4=a(7/2)^2+b(7/2)}}}
{{{4=(49/4)a+(7/2)b}}}
1.{{{49a+14b=16}}}
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.
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(7,0)
{{{0=a(7)^2+b(7)}}}
2.{{{49a+7b=0}}}
Subtract eq. 2 from eq. 1,
{{{49a+14b-49a-7b=16-0}}}
{{{7b=16}}}
{{{highlight(b=16/7)}}}
Then,
{{{49a+7(16/7)=0}}}
{{{49a+16=0}}}
{{{49a=-16}}}
{{{highlight(a=-(16/49))}}}
So then,
{{{y=-(16/49)x^2+(16/7)x}}}