# The line has equation y=2x+c and a curve has equation y=8-2x-x^2.

1) for the case where the line is a tangent to the curve, find the value of the constant c.
2) For the case where c = 11, find the x-coordinates of the points of intersection of the line and the curve. Find also, by integration, the area of region between the line and the curve.

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1. to solve:
2x+c = 8-2x-x^2
x^2 + 4x + c-8 = 0
to have y = 2x+c to be a tangent, there can be only one real root of the quadratic

so b^2 - 4ac = 0
16 - 4(1)(c-8) = 0
16 - 4c + 32 = 0
4c = 48
c = 48/4 = 12

check with Wolfram:
http://www.wolframalpha.com/input/?i=y%3D2x%2B12,+y%3D8-2x-x%5E2

for your second part
y = 2x+11, y = 8-2x - x^2

I will let you solve it to show:
http://www.wolframalpha.com/input/?i=y%3D2x%2B11,+y%3D8-2x-x%5E2

so for the area:
A = ∫ (8-2x-x^2 - 2x - 11) dx from -3 to -1
simplify and finish it

this looks very straightforward, I trust you can handle it

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2. No it isnt that straightforward u must take into account the area of the trapezium as well so the best thing to do is to draw out the diagram

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