How do you find the equation of the tangent line to curve #y=(x-1)/(x+1)#, that are parallel to the line #x-2y=2#?

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Answer 1 · 2018-04-29T16:31:47

There are two such lines:

#y = 1/2x - 1/2#
#y = 1/2x + 7/2#

We start by noticing that the line can be converted to

#x - 2 = 2y -> y = 1/2x - 1#

This line has a slope of #1/2#, and since the tangent line must have the same slope, the derivative is going to have to equal #1/2#.

We can find the derivative via the quotient rule.

#y' = (x + 1 - (x- 1))/(x +1)^2#

#y' = 2/(x+ 1)^2#

Set this to #1/2# and solve:

#1/2 = 2/(x +1)^2#

#(x +1)^2 = 4#

#x + 1 = +- 2#

#x= 1 or -3#

The corresponding values of the function at these points are

#y(1) = (1- 1)/(1 + 1) = 0#
#y(-3) = (-3 - 1)/(-3 + 1) = 2#

Now we must use find the tangent line equations.

#y - 0 = 1/2(x - 1) -> y = 1/2x - 1/2#

#y - 2 = 1/2(x +3) -> y = 1/2x + 7/2#

Our answer is clearly viable which can be proved graphically.

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Hopefully this helps!