Solution:

Step 1: Find where the denominator equals zero. $x^2 - 9 = 0$ $(x - 3)(x + 3) = 0$ $x = 3 \text{ or } x = -3$

Step 2: Determine the domain.

Domain: All real numbers except $x = 3$ and $x = -3$

In interval notation: $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$

Step 3: Check if numerator is zero at these points.

• At $x = 3$: $2(3) + 1 = 7 \neq 0$ ✓
• At $x = -3$: $2(-3) + 1 = -5 \neq 0$ ✓

Since the numerator is not zero at these points, both are vertical asymptotes.

Answer:

• Domain: $x \in \mathbb{R}, x \neq \pm 3$
• Vertical asymptotes: $x = 3$ and $x = -3$

Solution:

Part (a): Vertical asymptotes occur where the denominator equals zero (and the numerator doesn't).

Factor the denominator: $x^2 - 4x - 5 = (x - 5)(x + 1)$

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.

Solution:

Part a) $f(x) = \frac{3x^2 - 2x + 1}{5x^2 + 4}$

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

Solution:

Part (a): Factor both numerator and denominator:

Numerator: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ (difference of cubes) Denominator: $x^2 - 4 = (x - 2)(x + 2)$

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

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: $(0, 2)$

Solution:

Step 1: Factor the numerator and denominator completely.

Numerator: $x^2 - 4 = (x - 2)(x + 2)$

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 $x = 3$