SOLUTION: Find the equation of the tangent to the curve x^2y-x=y^3-8 at x=0.

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Question 1153034: Find the equation of the tangent to the curve x^2y-x=y^3-8 at x=0.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find the equation of the tangent to the curve x^2y-x=y^3-8 at x=0.
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I did this, but lost it before it was posted.
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@ x = 0, y = 2
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Differentiate implicitly and find the slope.
Slope = -1/12
---> y = (-x/12) + 2
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email via the TY note if you need help.

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
x^2y-x=y^3-8 at x=0
Substituting x=0,






So the point of tangency is (0,2), and we want the equation of the
green line below:



We find the slope of the tangent line, which is the
same as the derivative at that point:  So we find
the derivative implicitly, i.e., without solving for
the independent variable y:




We substitute x=0 and y=2 and solve for 









That's the slope of the tangent line at (0,2), which is
the green line. So



Since the point of tangency is the y-intercept (0,2), we can just
use:



with 




Edwin

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