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 $y = -x +5$