Question 934065
The slope of the tangent line to the curve is equal to the derivative at that point. 
{{{y=-x^2-2x+8}}}
{{{dy/dx=-2x-2}}}
So then set this value equal to the slope of the tangent line,
{{{-2x-2=2}}}
{{{-2x=4}}}
{{{x=-2}}}
At that point, both y values are the same.
{{{2x+c=-x^2-2x+8}}}
{{{2(-2)+c=-(-2)^2-2(-2)+8}}}
{{{-2+c=-4+4+8}}}
{{{c=10}}}
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{{{2x+11=-x^2-2x+8}}}
{{{x^2+4x+3=0}}}
{{{(x+1)(x+3)=0}}}
Two solutions,
{{{x+1=0}}}
{{{x=-1}}}
and
{{{x+3=0}}}
{{{x=-3}}}
Integrate between {{{x=-3}}} and {{{x=-1}}}
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{{{drawing(300,300,-6,6,-2,10,grid(1),circle(-3,5,0.3),circle(-1,9,0.3),graph(300,300,-6,6,-2,10,8-2x-x^2,2x+11))}}}
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{{{int(((8-2x-x^2)-(2x+11)),dx,-3,-1)=int(-(x^2+4x+3),dx,-3,-1)}}}
{{{int(((8-2x-x^2)-(2x+11)),dx,-3,-1)=-(x^3/3+2x^2+3x)+C}}}
{{{int(((8-2x-x^2)-(2x+11)),dx,-3,-1)=-((-1)^3/3+2(-1)^2+3(-1))+((-3)^3/3+2(-3)^2+3(-3))}}}
{{{int(((8-2x-x^2)-(2x+11)),dx,-3,-1)=-(-1/3+2-3)+(-9+18-9)}}}
{{{int(((8-2x-x^2)-(2x+11)),dx,-3,-1)=4/3}}}