Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 5

Determine the equation of the tangent line to the curve $\displaystyle y = \frac{x - 1}{x - 2}$ at the point $(3, 2)$

Using the definition $\text{(Slope of the tangent line)}$

We have $a = 3$ and $\displaystyle f(x) = \frac{x - 1}{x - 2}$, so the slope is


$\begin{equation} \begin{aligned} \displaystyle m &= \lim \limits_{x \to 3} \frac{f(x) - f(3)}{x - 3}\\ \\ \displaystyle m &= \lim \limits_{x \to 3} \frac{\frac{x - 1}{x - 2} - \left( \frac{3 - 1}{3 - 2} \right)}{x - 3} && \text{ Substitute value of $a$ and $x$}\\ \\ \displaystyle m &= \lim \limits_{x \to 3} \frac{\frac{x - 1}{x - 2} - 2}{x - 3} && \text{ Get the LCD and simplify}\\ \\ \displaystyle m &= \lim \limits_{x \to 3} \frac{-x + 3}{(x - 3)(x - 2)} && \text{ Factor the numerator}\\ \\ \displaystyle m &= \lim \limits_{x \to 3} \frac{-1 \cancel{(x - 3)}}{\cancel{(x - 3)}(x - 2)} && \text{ Cancel out like terms}\\ \\ \displaystyle m &= \lim \limits_{x \to 3} \frac{-1}{x - 2} = \frac{-1}{3-2} = -\frac{1}{1} = -1 && \text{ Evaluate the limit}\\ \end{aligned} \end{equation}$

Therefore,
The slope of the tangent line is $m = -1$
Using point slope form


$\begin{equation} \begin{aligned} y - y_1 =& m ( x - x_1)\\ \\ y - 2 =& -1 ( x - 3) && \text{ Substitute value of $x, y$ and $m$}\\ \\ y - 2 =& - x + 3 && \text{ Combine like terms}\\ \\ y =& -x+5 \end{aligned} \end{equation}$


Therefore,
The equation of the tangent line at $(3,2)$ is $y = -x +5$