Example 1: Rational Function with Linear Terms \( f(x) = \dfrac{x + 1}{x - 1} \)
Problem: Let \( f \) be a rational function defined by \( f(x) = \dfrac{x + 1}{x - 1} \).
- Find the domain of \( f \).
- Find the \( x \)- and \( y \)-intercepts of the graph of \( f \).
- Find the vertical and horizontal asymptotes for the graph of \( f \) if there are any.
- Use your answers to sketch the graph of \( f \).
Solution:
- Domain: All real numbers except \( x = 1 \), since this value makes the denominator zero.
-
Intercepts:
* \( x \)-intercept: Solve \( x + 1 = 0 \implies x = -1 \), yielding point \((-1, 0)\).
* \( y \)-intercept: Evaluate \( f(0) = \frac{0 + 1}{0 - 1} = -1 \), yielding point \((0, -1)\). -
Asymptotes:
* Vertical asymptote: Set denominator \( x - 1 = 0 \implies x = 1 \).
* Horizontal asymptote: Degrees of numerator and denominator are both equal to \( 1 \), so \( y = \frac{1}{1} = 1 \). -
Sign Chart & Sketching:
The zeros of numerator and denominator (\(-1\) and \(1\)) split the number line into intervals: \( (-\infty, -1) \), \((-1, 1)\), and \((1, +\infty)\).\( x \) \(-\infty\) \(-1\) \(1\) \(+\infty\) \( f(x) \) \(+\) \(0\) (x-intercept) \(-\) V.A. \(+\) * For \( x \in (-\infty, -1) \), \( f(x) > 0 \) and approaches \( y = 1 \) on the left.
* For \( x \in (-1, 1) \), \( f(x) < 0 \), passes through \((0, -1)\), and decreases without bound near \( x = 1 \).
* For \( x \in (1, +\infty) \), \( f(x) > 0 \), increases without bound from \( x = 1 \), and approaches \( y = 1 \) on the right.
