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

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To find the equation of the tangent line to the curve y = (x - 1)/(x - 2) at the point (3, 2), the slope m is calculated using the limit definition of the derivative. The limit simplifies to m = -1. Using the point-slope form of the line equation, the tangent line is determined to be y = -x + 5. A correction is noted regarding the sign in the final answer. The discussion emphasizes the importance of careful algebraic manipulation in derivative calculations.
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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.
 
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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
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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