SOLUTION: a line has the equation y=2x+c and a curve has the equation y= 8-2x-x^2 the line is tangent to the curve •find the value constant to c •for the case where c=11 find the x-

Algebra ->  Trigonometry-basics -> SOLUTION: a line has the equation y=2x+c and a curve has the equation y= 8-2x-x^2 the line is tangent to the curve •find the value constant to c •for the case where c=11 find the x-      Log On


   

Question 934065: a line has the equation y=2x+c and a curve has the equation y= 8-2x-x^2
the line is tangent to the curve
•find the value constant to c
•for the case where c=11 find the x-coordinates of points of intersection of the line and the curve. Find by integration, the area of the region between the line and the curve
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The slope of the tangent line to the curve is equal to the derivative at that point.
y=-x%5E2-2x%2B8
dy%2Fdx=-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%2Bc=-x%5E2-2x%2B8
2%28-2%29%2Bc=-%28-2%29%5E2-2%28-2%29%2B8
-2%2Bc=-4%2B4%2B8
c=10
.
.
.
2x%2B11=-x%5E2-2x%2B8
x%5E2%2B4x%2B3=0
%28x%2B1%29%28x%2B3%29=0
Two solutions,
x%2B1=0
x=-1
and
x%2B3=0
x=-3
Integrate between x=-3 and x=-1
.
.
.

.
.
.




int%28%28%288-2x-x%5E2%29-%282x%2B11%29%29%2Cdx%2C-3%2C-1%29=4%2F3