Equation of a Tangent to a Curve

**bearhug** · March 4th 2009, 02:36 PM

Find the points on the curve y = 2 / (3x-2) where the tangent is parallel to the line y = (-3/2)x-1

I’m not sure I understand what they are asking me to do and how I go about answering this question. Any help is appreciated.

**skeeter** · March 4th 2009, 02:40 PM

Originally Posted by bearhug

Find the points on the curve y = 2 / (3x-2) where the tangent is parallel to the line y = (-3/2)x-1

I’m not sure I understand what they are asking me to do and how I go about answering this question. Any help is appreciated.

$y = \frac{2}{3x-2}$

1. find the derivative of the above function.

2. set the derivative equal to $-\frac{3}{2}$ ... (why do that ?)

3. solve for x

4. determine the point(s), (x,y), on the curve using the x-value(s) found in step 3.

**bearhug** · March 4th 2009, 02:46 PM

Because -3/2 is the slope
Then I find the x-coordinate
and using the x-coordinate I find the point on the y-coordinate....then I will have my point?

**bearhug** · March 4th 2009, 02:53 PM

Is the derivative:

$<br /> y = \frac{-1}{(x-2)^2}<br />$
??

**skeeter** · March 4th 2009, 02:59 PM

familiar with the chain rule?

$y = \frac{2}{3x-2} = 2(3x-2)^{-1}$

$y' = -2(3x-2)^{-2} \cdot 3<br />$

$y' = -\frac{6}{(3x-2)^2}$

finish.

**bearhug** · March 4th 2009, 03:24 PM

Sorry about that, must be tired from all the math.

When I sub the slope (-3/2) into the derivative I get -0.14201
That seems wrong.....is it?

**skeeter** · March 4th 2009, 03:44 PM

you do not substitute $-\frac{3}{2}$ in for $x$ ... you set the derivative equal to $-\frac{3}{2}$ and then solve for $x$ , remember?

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