where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$.

Example: $f(x) = \frac{2x^2 - 3x + 1}{x^2 - 4}$

Domain of Rational Functions

The domain includes all real numbers except where the denominator equals zero.

To find the domain:

1. Set the denominator $Q(x) = 0$
2. Solve for $x$
3. Domain is all real numbers except these values

Vertical Asymptotes

A vertical asymptote occurs at values of $x$ where the denominator is zero but the numerator is not zero.

Finding Vertical Asymptotes

1. Factor both numerator and denominator completely
2. Cancel any common factors (these create holes, not asymptotes)
3. Set remaining denominator factors equal to zero
4. The solutions are the vertical asymptotes

Example: $f(x) = \frac{x + 1}{x - 3}$ has a vertical asymptote at $x = 3$

Horizontal Asymptotes

A horizontal asymptote describes the end behavior of the function as $x \to \pm\infty$.

Finding Horizontal Asymptotes

Compare the degrees of the numerator and denominator:

1. If degree of numerator < degree of denominator:

   • Horizontal asymptote: $y = 0$

2. If degree of numerator = degree of denominator:

   • Horizontal asymptote: $y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$

3. If degree of numerator > degree of denominator:

   • No horizontal asymptote (there may be a slant/oblique asymptote)

Holes (Removable Discontinuities)

A hole occurs when there is a common factor in both numerator and denominator.

Finding Holes

1. Factor completely
2. Identify common factors
3. Cancel the common factors
4. The zero of the canceled factor gives the x-coordinate of the hole
5. Substitute this x-value into the simplified function to get the y-coordinate

Summary Table

Key Difference

• Vertical Asymptote: Function approaches $\pm\infty$
• Hole: Function is undefined but could be "filled in"

Factor the numerator: $2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2)$

So: $f(x) = \frac{2(x - 2)(x + 2)}{(x - 5)(x + 1)}$

The denominator is zero when $x = 5$ or $x = -1$.

Since neither of these values makes the numerator zero, we have vertical asymptotes at $x = 5$ and $x = -1$.

Part (b): For horizontal asymptotes, compare the degrees of numerator and denominator.

Both have degree 2, so we take the ratio of leading coefficients:

$y = \frac{2}{1} = 2$

Horizontal asymptote: $y = 2$

Part (c): Holes occur when a factor cancels from both numerator and denominator.

Looking at our factored form, there are no common factors, so there are no holes.

Degree of numerator = 2 Degree of denominator = 2 Degrees are equal.

Horizontal asymptote: $y = \frac{3}{5}$ (ratio of leading coefficients)

Part b) $g(x) = \frac{2x + 5}{x^3 - 1}$

Degree of numerator = 1 Degree of denominator = 3 Numerator degree less than denominator degree.

Horizontal asymptote: $y = 0$

Part c) $h(x) = \frac{4x^3 + 2x}{2x^2 - 3}$

Degree of numerator = 3 Degree of denominator = 2 Numerator degree greater than denominator degree.

No horizontal asymptote (there is a slant asymptote instead)

Answers:

• a) $y = \frac{3}{5}$
• b) $y = 0$
• c) No horizontal asymptote

$g(x) = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$

The factor $(x - 2)$ cancels, creating a hole at $x = 2$.

To find the $y$-coordinate of the hole, substitute into the simplified function:

$g(x) = \frac{x^2 + 2x + 4}{x + 2}$ (for $x \neq 2$)

At $x = 2$: $y = \frac{4 + 4 + 4}{2 + 2} = \frac{12}{4} = 3$

Hole at $(2, 3)$

Part (b):

Vertical asymptote: From the simplified denominator, $x + 2 = 0$ gives $x = -2$

Horizontal/Oblique asymptote: The simplified numerator has degree 2, denominator has degree 1.

Since numerator degree > denominator degree by exactly 1, we have an oblique asymptote.

Divide: $\frac{x^2 + 2x + 4}{x + 2}$

Using polynomial long division: $x^2 + 2x + 4 = (x + 2)(x) + 4$ Continue: $x + \frac{4}{x+2}$

Actually, let me redo this: $(x^2 + 2x + 4) \div (x + 2)$:

• $x^2 \div x = x$
• $x(x + 2) = x^2 + 2x$
• $(x^2 + 2x + 4) - (x^2 + 2x) = 4$
• $4 \div (x + 2)$ leaves remainder

So $g(x) = x + \frac{4}{x + 2}$

Wait, let me be more careful: $\frac{x^2 + 2x + 4}{x + 2} = \frac{x^2 + 2x}{x + 2} + \frac{4}{x + 2} = x + \frac{4}{x + 2}$

As $x \to \pm\infty$, the $\frac{4}{x+2} \to 0$, so oblique asymptote: $y = x$

Part (c): The $y$-intercept occurs at $x = 0$:

$g(0) = \frac{0 - 8}{0 - 4} = \frac{-8}{-4} = 2$

$y$-intercept: $(0, 2)$

Denominator: $x^2 - x - 6 = (x - 3)(x + 2)$

$f(x) = \frac{(x - 2)(x + 2)}{(x - 3)(x + 2)}$

Step 2: Identify common factors.

Common factor: $(x + 2)$

This means there is a hole at $x = -2$.

Step 3: Cancel the common factor.

$f(x) = \frac{x - 2}{x - 3}, \quad x \neq -2$

Step 4: Find vertical asymptotes from the simplified function.

Set denominator equal to zero: $x - 3 = 0$

Vertical asymptote at $x = 3$

Step 5: Find the y-coordinate of the hole.

Substitute $x = -2$ into the simplified function: $y = \frac{-2 - 2}{-2 - 3} = \frac{-4}{-5} = \frac{4}{5}$

Answer:

• Hole at $\left(-2, \frac{4}{5}\right)$
• Vertical asymptote at $x = 3$