SOLUTION: Find the equation of the tangent line to the curve {{{(x-y)^2}}}{{{""=""}}}{{{2x+1}}} at the point (4,1).

Algebra ->  Functions -> SOLUTION: Find the equation of the tangent line to the curve {{{(x-y)^2}}}{{{""=""}}}{{{2x+1}}} at the point (4,1).      Log On


   

Question 1003954: Find the equation of the tangent line to the curve %28x-y%29%5E2%22%22=%22%222x%2B1
at the point (4,1).
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of the tangent line to the curve
%28x-y%29%5E2%22%22=%22%222x%2B1 at the point (4,1).
%28x-y%29%5E2%22%22=%22%222x%2B1

We take the derivative implicitly:

2%28x-y%29%281-expr%28dy%2Fdx%29%29%22%22=%22%222

Substitute (x,y) = (4,1)

2%284-1%29%281-expr%28dy%2Fdx%29%29%22%22=%22%222

2%283%29%281-expr%28dy%2Fdx%29%29%22%22=%22%222

6%281-expr%28dy%2Fdx%29%29%22%22=%22%222

3%281-expr%28dy%2Fdx%29%29%22%22=%22%221

1-expr%28dy%2Fdx%29%22%22=%22%221%2F3

-dy%2Fdx%22%22=%22%221%2F3-1

-dy%2Fdx%22%22=%22%221%2F3-3%2F3

-dy%2Fdx%22%22=%22%22-2%2F3

dy%2Fdx%22%22=%22%222%2F3

The derivative at a point is the slope of the
line tangent to the curve at the point, therefore
the slope of the tangent line is 

m%22%22=%22%222%2F3

We use the point-slope formula for the equation
of the tangent line:

y-y%5B1%5D%22%22=%22%22m%28x-x%5B1%5D%29

where (x1,y1) = (4,1)

[Do not substitute for x and y, but leave them 
as variables]

y-1%22%22=%22%22expr%282%2F3%29%28x-4%29

y-1%22%22=%22%22expr%282%2F3%29x-8%2F3

y%22%22=%22%22expr%282%2F3%29x-8%2F3%2B1

y%22%22=%22%22expr%282%2F3%29x-8%2F3%2B3%2F3

y%22%22=%22%22expr%282%2F3%29x-5%2F3, graphed in green below:



Edwin