Question

identify the horizontal and vertical asymptotes. then, sketch the functions.

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

Explanation:

Problem 13: \( f(x) = \frac{1}{4x + 4} \)

Step 1: Find Vertical Asymptote

To find the vertical asymptote, we set the denominator equal to zero and solve for \( x \):
\( 4x + 4 = 0 \)
Subtract 4 from both sides: \( 4x = -4 \)
Divide by 4: \( x = -1 \)
So, the vertical asymptote is \( x = -1 \).

Step 2: Find Horizontal Asymptote

For a rational function \( f(x) = \frac{N(x)}{D(x)} \), if the degree of \( N(x) \) is less than the degree of \( D(x) \), the horizontal asymptote is \( y = 0 \).
Here, the degree of the numerator (1) is 0 (since it's a constant) and the degree of the denominator ( \( 4x + 4 \)) is 1. Since \( 0 < 1 \), the horizontal asymptote is \( y = 0 \).

Step 1: Factor the Denominator

First, factor the denominator \( x^2 - 2x - 3 \):
\( x^2 - 2x - 3 = (x - 3)(x + 1) \) (by finding two numbers that multiply to -3 and add to -2: -3 and 1)

Step 2: Find Vertical Asymptotes

Set the denominator equal to zero and solve for \( x \):
\( (x - 3)(x + 1) = 0 \)
This gives \( x - 3 = 0 \) or \( x + 1 = 0 \), so \( x = 3 \) or \( x = -1 \). These are the vertical asymptotes.

Step 3: Simplify the Function (Check for Holes)

The numerator is \( x - 1 \), and the denominator factors to \( (x - 3)(x + 1) \). There are no common factors between the numerator and the denominator, so there are no holes.

Step 4: Find Horizontal Asymptote

For a rational function \( f(x) = \frac{N(x)}{D(x)} \), if the degree of \( N(x) \) is less than the degree of \( D(x) \), the horizontal asymptote is \( y = 0 \).
Here, the degree of the numerator ( \( x - 1 \)) is 1, and the degree of the denominator ( \( x^2 - 2x - 3 \)) is 2. Since \( 1 < 2 \), the horizontal asymptote is \( y = 0 \).

Answer:

• Vertical Asymptote: \( x = -1 \)
• Horizontal Asymptote: \( y = 0 \)

Problem 14: \( f(x) = \frac{x - 1}{x^2 - 2x - 3} \)