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

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Gió Share
May 23, 2015

Have a look:
enter image source here
which gives yu two equations:
#y=1/2x+7/2#
and
#y=1/2x-1/2#

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Jim H Share
May 23, 2015

Find the equations of the tangent lines to the curve #y= (x-1)/(x+1)# that are parallel to the line #x-2y = 2#.

There is a bit of algebra and arithmetic for this. Let's focus on the reasoning and the calculus.

One
A line parallel to #x-2y = 2# must have the same slope. The slope of this line is #1/2#. So we want the slope of the tangent line to be #1/2#

Two
How do we find the slope of the tangent line? -- The derivative. So, we want the derivative to be #1/2#.

What is the derivative of #y= (x-1)/(x+1)# ?

Use the quotient rule:

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

Three

Find #x# to make #y'=1/2#

#2/(x+1)^2 = 1/2# if and only if

#(x+1)^2 =4#

So #x+1 = +-2#

and #x=1, -3#

Four

Find the #y# values at #x=1# (#y=0#)and at #x=-3 (#y=2)#

Five

Find the equations of the lines:

through #(1,0)# with slope #m=1/2#

and through #(-3,2)# with slope #m=1/2#.

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