ahhhh calc question?!? help??

Question

Find equations of the tangent lines to the curve below that are parallel to the line x - 2y = 2.

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

y = (smaller y-intercept)
y = (larger y-intercept)

thanks!

i got y=(-8/100)(x-3)+.3

Answer 1

Rewrite the line x - 2y = 2 in the form of y = mx + b to find the slope. Your tangent lines have to have the same slope.

Remember that the derivative of a function tells you the slope of the tangent line at that point. So take the derivative of the function, then set it equal to the slope you're looking for. Solve for x. This tells you the x intercept(s) of the tangent lines. Plug it back into the original equation to find the y value(s). Now that you have the slope and a point of the tangent line(s), you can find the equation for those tangent(s).

Answer 2

Putting
x - 2y = 2
into slope-intercept form,
y = (1/2)x - 1

For
y = (x-1)/(x+1)
y' = (x + 1 - x + 1) / (x + 1)^2 = 2 / (x + 1)^2
Setting y' = 1/2,
2 / (x + 1)^2 = 1/2
(x + 1)^2 = 4
(x + 1)^2 - 4 = 0
(x + 1 + 2)(x + 1 - 2) = 0
x = - 3, 1
y = (- 3 - 1) / (- 3 + 1), (1 - 1) / (1 + 1)
y = 2, 0
so the points of tangency are
(- 3, 2), (1, 0)
Line 1:
y - 2 = (1/2)(x - 3)
y = (1/2)x + 1/2
Line 2:
y - 1 = (1/2)(x - 0)
y = (1/2)x + 1