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I'm having trouble figuring out how to find the equation to a tangent line off a curve. I know you use implicit differentiation of the function, but lets say the derivative found is still a difficult function, such as $-2x + y^2/-2xy + 1$ Now, the formula for the tangent line is y = mx + c and lets say it passes through the point (-3, -5). How would you go about figuring this out?

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  • $\begingroup$ Is (-3,-5) the point of tangency? $\endgroup$ – S. Dolan Sep 7 at 7:45
  • $\begingroup$ Do you mean $\dfrac{-2x+y^2}{-2xy+1}$? $\endgroup$ – José Carlos Santos Sep 7 at 7:52
  • $\begingroup$ Yes I mean that $\endgroup$ – DuncanK3 Sep 8 at 4:03
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Firstly you find $\frac {dy}{dx}$. For this you only need to use implicit differentiation if you do not have an explicit formula for $y$. Then substitute the coordinates of the point of tangency into your expression for $\frac {dy}{dx}$.

Example $\frac {-2x+y^2}{-2xy+1}=\frac {-31}{29}$ at $(-3,-5)$.

This gradient is the $"m"$ in $y=mx+c$.

Example $y=\frac {-31}{29}x+c$. You then use the point $(-3,-5)$ to find $"c"$.

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  • $\begingroup$ So all I have to do is sub in the x and y values to find the gradient, m? $\endgroup$ – DuncanK3 Sep 8 at 4:04
  • $\begingroup$ That's right. At a point of tangency the gradient of the curve is the same as the gradient $m$ of the line. $\endgroup$ – S. Dolan Sep 8 at 7:41

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