To analyze the rational function \( f(x) = \frac{x^2 + 14x + 45}{3x^2 + 27x} \), we will determine the vertical asymptotes, horizontal asymptotes, and any holes.
Step 1: Factor the Function
Let's start by factoring the numerator and the denominator.
Numerator: \[ x^2 + 14x + 45 = (x + 5)(x + 9) \]
Denominator: \[ 3x^2 + 27x = 3x(x + 9) \]
Step 2: Identify Holes
A hole occurs when there is a common factor in both the numerator and denominator. In this case, both the numerator and denominator have a factor of \( (x + 9) \).
Thus, there is a hole at \( x = -9 \).
Step 3: Identify Vertical Asymptotes
Vertical asymptotes occur where the denominator is zero (and not canceled from the numerator). After canceling the common factor \( (x + 9) \):
The simplified form is: \[ f(x) = \frac{x + 5}{3x} \]
Now, set the denominator equal to zero: \[ 3x = 0 \] \[ x = 0 \]
So, there is a vertical asymptote at \( x = 0 \).
Step 4: Identify Horizontal Asymptotes
To find horizontal asymptotes in rational functions, we compare the degrees of the numerator and the denominator.
Both the numerator and denominator are degree 2 when considering the original function. When the degrees are equal, the horizontal asymptote is found by taking the ratio of the leading coefficients.
The leading coefficient of the numerator is \( 1 \) and the leading coefficient of the denominator is \( 3 \).
Thus, the horizontal asymptote is: \[ y = \frac{1}{3} \]
Summary of Characteristics
- Hole: \( x = -9 \)
- Vertical asymptote: \( x = 0 \)
- Horizontal asymptote: \( y = \frac{1}{3} \)
Answer
- There is a vertical asymptote at \( x = 0 \).
- There is a horizontal asymptote at \( y = \frac{1}{3} \).
- There is a hole at \( x = -9 \).