# Finding equation of derivative that is parallel to another line

1. Oct 10, 2008

### hachi_roku

1. The problem statement, all variables and given/known data
find equations of the tangent lines to the curve
y= x-1/x+1

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

2. Relevant equations

3. 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!

2. Oct 10, 2008

### gabbagabbahey

Are you sure that the slope of x-2y=2 is -1? ;0)

3. Oct 10, 2008

### hachi_roku

oops

4. Oct 10, 2008

### hachi_roku

so the slope is 1/2.....i plugg that in so my answer is now y = x-1. is this correct?

5. Oct 10, 2008

### gabbagabbahey

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?

6. Oct 10, 2008

### hachi_roku

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

7. Oct 10, 2008

### gabbagabbahey

your dy/dx is right; but what are the solutions to x^2+2x-3=0?

8. Oct 10, 2008

### hachi_roku

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

9. Oct 10, 2008

### gabbagabbahey

how do you get x=-2 or -4/2?! (-2)^2+2(-2)-3=-3 not zero...did you try factoring the quadratic?

10. Oct 10, 2008

### hachi_roku

i meant -2+or-4 all over 2.

11. Oct 10, 2008

### gabbagabbahey

Either way its still incorrect; x^2+2x-3=(x+3)(x-1) so x=1 or x=-3.

12. Oct 10, 2008

### hachi_roku

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. Oct 10, 2008

### gabbagabbahey

Okay, looks good. :0)