How Do You Find the Equation of a Tangent Line to a Curve?

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SUMMARY

The discussion focuses on finding the equation of the tangent line to the curve defined by the function y = (x - 1)/(x - 2) at the point (3, 2). The derivative is calculated using the limit definition of the derivative, resulting in a slope of -1. Consequently, the equation of the tangent line is determined to be y = -x + 5. Participants emphasize the importance of correctly applying the limit process and adjusting signs in the final equation.

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Homework Statement



Find an equation of the tangent line to the curve at the given point.

y = \frac{x - 1}{x - 2} , (3, 2)

Homework Equations



m = \lim_{h\to 0} \frac{f(a + h) - f(a)}{h}

y = mx + b

The Attempt at a Solution



\lim_{h\to 0} \frac{\frac{(x + h) - 1}{(x + h) - 2} - \frac{x - 1}{x - 2}}{h}

\lim_{h\to 0} \frac{\frac{2 + h}{1 + h} - \frac{2}{1}}{h}

\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2}{h}

\lim_{h\to 0} \frac{2 + h}{h + h^2} - \frac{2 + 2h}{h + h^2}

\lim_{h\to 0} \frac{h + 2h}{h + h^2}I'm not sure what to do after this step. The final answer is y = -x + 5
Thanks.
 
Last edited:
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Your last line is wrong (remember to change + to -!).
We have:
\lim_{h\to{0}}(\frac{2+h}{h(1+h)}-\frac{2}{h})=\lim_{h\to{0}}\frac{2+h-2(1+h)}{h(1+h)}=\lim_{h\to{0}}\frac{-h}{h(1+h)}=\lim_{h\to{0}}\frac{-1}{(1+h)}=-1
 

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