Finding equation of derivative that is parallel to another line

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Homework Help Overview

The discussion revolves around finding the equations of tangent lines to the curve defined by y = (x - 1)/(x + 1) that are parallel to the line represented by the equation x - 2y = 2. Participants are exploring the relationship between the derivative of the curve and the slope of the given line.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss finding the derivative of the curve and setting it equal to the slope of the line to determine the tangent lines. There are questions about the correctness of the slope and the resulting equations. Some participants express uncertainty about their calculations and seek clarification on the solutions to the quadratic equation derived from the derivative.

Discussion Status

The discussion is active, with participants correcting each other's calculations and confirming the slope of the line. There is a focus on deriving the correct x-values from the quadratic equation, and some participants have provided guidance on factoring and solving the equation. Multiple interpretations of the problem are being explored, particularly regarding the tangent lines.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can provide. There is an ongoing examination of assumptions related to the slope and the setup of the problem.

hachi_roku
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Homework Statement


find equations of the tangent lines to the curve
y= x-1/x+1


that are parallel to the line x-2y = 2


Homework Equations





The Attempt at a Solution


ok so i found the derivative, then set it equal to the slope (-1). then plugged x back into the original equation to get

y = -x + 41/14 and y= -x + 17/10


is this right? thanks in advance!
 
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Are you sure that the slope of x-2y=2 is -1? ;0)
 
oops
 
so the slope is 1/2...i plugg that in so my answer is now y = x-1. is this correct?
 
Your slope is right now, but you should be getting two lines; and the one you have is incorrect. Let's star with your solutions for x from setting dy/dx=1/2...what do you get for those?
 
from there, my solutions for x = x^2 + 2X + 1 = 4...then x^2 + 2x -3 = 0.

is my dy/dx wrong? i get 2/(x+1)^2
 
your dy/dx is right; but what are the solutions to x^2+2x-3=0?
 
my solutions are x = -2+or-4/2. from there one equation i get is y= x/2+7/2 and the other is y=x/2-1/2
 
how do you get x=-2 or -4/2?! (-2)^2+2(-2)-3=-3 not zero...did you try factoring the quadratic?
 
  • #10
i meant -2+or-4 all over 2.
 
  • #11
Either way its still incorrect; x^2+2x-3=(x+3)(x-1) so x=1 or x=-3.
 
  • #12
ya tahts what i got...i was tryin to say that x = -2+-4/2 which equaled 1,-3. my mistake.

when i plug them in and put in the form i get y=x/2+7/2 and y= x/2-1/2. thanks for your help btw
 
  • #13
Okay, looks good. :0)
 

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