Your derivative is wrong . . .
Find the points on the curve
where the tangent is parallel to the line
I know the slope of the tangent is .
I can express the curve as: .
The derivative is: .
So we have: .
Find the points on the curve y = 2/3x + 2 where the tangent is parallel to the line y = (-3/2x) - 1
I know the slope of the tangent is -3/2 as it is parallel to the line y = (-3/2x) - 1.
Therefore m = -3/2
So I can express the curve as...
y = 2(3x - 2)^-1
Using a combination of the product rule and chain rule I have come up with:
= -6/(3x - 2)^2 + 1/(3x - 2)
-3/2 = -6/(3x - 2)^2 + 1/(3x - 2)
But that's where I get hung up. Can I express that as the following then solve for x?
-3/2 = -5/(3x - 2)^2 + (3x - 2)
Oh my, it seems to me like you're complicating it.
The way I see it, you're looking at a spot on the derivative of y where y' = -3/2.
Now, the derivative of a constant is 0, so the 2nd term is 0 now.
Also, remember that 2/3 is also a constant and can thus be taken out of the d/dx.
I trust you can solve this? (I don't know how to do the plus or minus thing with LaTex)
dy/dx means taking the derivative of y with respect to x, or differentiating y with x as the variable of differentiation.
But remember, though y = 2/3x - 2, it is still y = 2(3x-2)^-1
y' = -6(3x-2)^-2
Then plug in y = -3/2
(3x-2)^2 = 4
3x - 2 = 2 or 3x - 2 = -2
x = 4/3 or x = 0
And please use parentheses to clarify what you mean!
You talk about the line "y = (-3/2x) - 1" so that is clearly supposed to be (-3/2)x but you also have "y = 2/3x + 2" which we clearly are supposed to interpret as y= 2/(3x)+ 2 because if we interpret it the same way, as y= (2/3)x+ 2, is a line with slope 2/3 and the two lines are perpendicular.