Question 999156
Here's the graph.
Any guess is a good one as long as the {{{x[v]>8}}} and {{{y[v]>7}}}.
*[illustration v6.JPG].
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Using the model,
{{{y=ax^2+bx+c}}}
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{{{1=a(0)^2+b(0)+c}}}
1.{{{c=1}}}
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{{{5=a(4)^2+b(4)+1}}}
{{{16a+4b=4}}}
2.{{{4a+b=1}}}
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{{{7=a(8)^2+b(8)+1}}}
{{{64a+8b=6}}}
3.{{{32a+4b=3}}}
From eq. 2,
{{{4a+b=1}}}
{{{16a+4b=4}}}
Subtracting from eq. 3,
{{{32a+4b-(16a+4b)=3-4}}}
{{{16a=-1}}}
{{{a=-1/16}}}
Then,
{{{16(-1/16)+4b=4}}}
{{{-1+4b=4}}}
{{{4b=5}}}
{{{b=5/4}}}
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{{{y=-x^2/16+(5/4)x+1}}}
{{{y=-(1/16)(x^2-20x)+1}}}
{{{y=-(1/16)(x^2-20x+100)+1+(1/16)(100)}}}
{{{y=-(1/16)(x-10)^2+4/4+25/4}}}
{{{y=-(1/16)(x-10)^2+29/4}}}
Vertex is ({{{10}}},{{{29/4}}}).
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*[illustration v7.JPG].