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.
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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. 1. find the derivative of the above function. 2. set the derivative equal to ... (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.
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?
Is the derivative: ??
familiar with the chain rule? finish.
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?
you do not substitute in for ... you set the derivative equal to and then solve for , remember?
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