Calculus_split

[deriviator]

Find the points on the curve y=(2)/(3x-2) where the tangent is paralle to the line y=-(3)/(2)x-1.
I need help with finding the derivative and the points.

y= (2)/(3x-2)
= 2(3x-2)^-2
[=2(3x-2)^-2' +(2)'(3x-2)^-2 or 2(3x)+(?)(3x-2)^-2] this step I'm not too sure
=-6(3x-2)^-2

y'= -6(3x-2)^-2
-(3)/(2)x= -6(3x-2)^-2
x=?

y=(2)/(3(?)-2)
=?

 

[Deleted member 4993]

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

First - start a new thread with a new problem

Second - read "Read before Posting" - show your work.

Hint to start the problem.

What is the relationship between "the slope of the tangent line at a point on the curve" and "the derivative of the function depicting the curve"?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[mmm4444bot]

Step 1: Find the slope of the line y = -(3/2)x - 1