Question 1056385
{{{du/dt=pt+c}}}
{{{u=(pt^2)/2+ct+d}}} where {{{d}}} is a constant.
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In vertex form,
{{{u=a(t-1)^2+5.5}}}
{{{u=a(t^2-2t+1)+5.5}}}
{{{u=at^2-2at+(a+5.5)}}}
Comparing terms,
{{{a=p/2}}}
{{{c=-2a=-p}}}
{{{d=a+5.5}}}
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Using the rate of change,
{{{-3=p(2)+c}}}
{{{2p+c=-3}}}
{{{2(-c)+c=-3}}}
{{{-2c+c=-3}}}
{{{-c=-3}}}
{{{c=3}}}
So then,
{{{p=-3}}}
{{{a=-3/2}}}
and
{{{d=-3/2+11/2}}}
{{{d=8/2}}}
{{{d=4}}}
So,
{{{u=-(3/2)(t-1)^2+5.5}}}