Find the equations of the tangent lines to the graph of the equation below that are parallel to the line given?

Question

f(x)=(x+1)/(x-1) 

2y+x=6 

y=? 

y=?

Answer 1

The function is f ( x ) = (x+1)/(x-1)

Using quotient rule differentiation d/dx(uv) = (vdu/dx-udv/dx)/v^2

f  ' (x ) = ((x-1)d/dx(x+1)-(x+1)d/dx(x-1))/(x-1)^2

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

           =-2/(x-1)^2

This is the slope of the tangent line .

Convert the line equation to slope - intercept form:

2y +x=6

2y=-x+6

y=(-1/2)x+3

Compare the given line y=(-1/2)x+3 with slope - intercept form of a line equation is y = mx + b, where m is slope, and b is y - intercept.

Slope (m ) = - 1/2.

From the given data, the tangent line is parallel to y=(-1/2)x+3.

Parallel lines have the same slope.

So,

-2/(x-1)^2= - 1/2

(x-1) ^2 = 4

x-1 = ±2

x-1 = 2.

x = 3.

x-1 = -2.

x = -1

Substitute x=3 in curve equation.

y = (x+1)/(x-1) =(3+1)/(3-1) =2

Substitute x=-1 in curve equation.

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

The points on curve is (3,2)(-1,0).

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The tangent equation parallel to y=(-1/2)x+3 is y =(-1/2)x+b

At (-1,0)

0 =(-1/2)(-1)+b

b = 0-(1/2)=-1/2

So tangent equation is y =(-1/2)x-1/2

At (3,2)

2 =(-1/2)(3)+b

b = 2+(3/2)=7/2

So tangent equation is y =(-1/2)x+7/2