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#?

Question

Cameron Alexander · Accepted Answer

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#.

Emily Ammerman · Answer

Have a look:

Evelyn Agard · Answer

undefined To find the equations of the tangent lines to the curve y = (x-1)/(x+1) that are parallel to the line x-2y = 2, we need to find the derivative of the curve and set it equal to the slope of the given line.

First, find the derivative of y with respect to x by using the quotient rule:

dy/dx = [(x+1)(1) - (x-1)(1)] / (x+1)^2

Next, set the derivative equal to the slope of the given line, which is 1/2 (since the line is in the form x-2y = 2):

[(x+1)(1) - (x-1)(1)] / (x+1)^2 = 1/2

Simplify the equation:

2[(x+1) - (x-1)] = (x+1)^2

Expand and simplify further:

2x + 2 - 2x + 2 = x^2 + 2x + 1

Combine like terms:

4 = x^2 + 2x + 1

Rearrange the equation:

x^2 + 2x - 3 = 0

Factor the quadratic equation:

(x+3)(x-1) = 0

Solve for x:

x = -3 or x = 1

Now that we have the x-values, substitute them back into the original equation y = (x-1)/(x+1) to find the corresponding y-values:

For x = -3, y = (-3-1)/(-3+1) = -4/-2 = 2

For x = 1, y = (1-1)/(1+1) = 0/2 = 0

So, the two points on the curve are (-3, 2) and (1, 0).

Finally, use the point-slope form of a line to find the equations of the tangent lines:

For the point (-3, 2), the equation of the tangent line is y - 2 = (1/2)(x + 3)

For the point (1, 0), the equation of the tangent line is y - 0 = (1/2)(x - 1)