The question is: Find the points to y = 2/(3x-2) where the tangent is parallel to the line y = (-3/2)x - 1.

I have gotten this far: (-3/2) = -6/(3x-2)^2

Where do I go from there?(Headbang)

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- Jan 5th 2010, 04:13 PMCuriousThoughtTangent Parallel to a Line Question
The question is: Find the points to y = 2/(3x-2) where the tangent is parallel to the line y = (-3/2)x - 1.

I have gotten this far: (-3/2) = -6/(3x-2)^2

Where do I go from there?(Headbang) - Jan 5th 2010, 04:20 PMHallsofIvy
Please, just use "regular" print! Because most internet readers do not respect "spaces", I cannot read your functions.

- Jan 5th 2010, 04:24 PMCuriousThought
Done. Apologies. I tried and failed to make it easier for you. Sorry.(Wink)

- Jan 5th 2010, 04:26 PMbigwave
see next reply

- Jan 5th 2010, 04:29 PMCuriousThought
- Jan 5th 2010, 04:49 PMHallsofIvy
Excellent! Okay, the derivative of y= 2/(3x-2)= 2(3x-2)^(-1) is y'= -2(3x-2)^(-2)(3)= -6(3x-2)^(-2). The slope of the tangent line at each point is equal to that. In order that the tangent line be parallel to y= (-3/2)x- 1, which has slope -3/2, we must have -6(3x-2)^(-2)= -3/2. That's what you have. Good! To solve that for x, first get rid of the fractions by multiplying on both sides by 2(3x-2)^2 and dividing by -3. You should get 4= (3x-2)^2. Taking the square root of both sides, $\displaystyle 3x-2= \pm 2$. The positive sign gives 3x= 4 so x= 4/3. The negative sign gives x= 0. When x= 4/3, y= 2/(3(4/3)- 2)= 2/2= 1 and when x= 0, y= 2/(3(0)-2)= -1. Thus the tangent line to the curve will be parallel to y= (-3/2)x- 1 at (4/3, 1) and (0, -1). You should be able to determine the "b" in y= (-3/2)x+ b for both of those points.

- Jan 5th 2010, 06:05 PMCuriousThought
Quick question: If I were to add (-3/2) to both sides, I would be left with 0= -6/(3x -2)^2 +3/2. Doing the resulting work should yield this: 27x^2 - 36x/(18x^2 - 24x +8) = 0. Why couldn't I multiply both sides by (18x^2 - 24x + 8)? I apologize if this is a juvenile question...(Wondering)

P.S. Thank you for what was a seemingly easy solution. I cannot believe I did not see that....D'oh! - Jan 5th 2010, 06:40 PMHallsofIvy
- Jan 5th 2010, 06:57 PMCuriousThought
Thank's again. (Wink)