Question

Find the vertical, horizontal, and oblique asymptotes, if any, of each rational function. $$ Q(x)=\frac{2 x^{2}-5 x-12}{3 x^{2}-11 x-4} $$

Short answer

Vertical asymptote: \( x = -\frac{1}{3} \). Horizontal asymptote: \( y = \frac{2}{3} \). No oblique asymptote.

Step-by-step solution

- Factor the numerator and the denominator

Factor both the numerator and the denominator of the rational function if possible. For the given function \( Q(x) = \frac{2x^2 - 5x - 12}{3x^2 - 11x - 4} \), factor each polynomial: Numerator: \( 2x^2 - 5x - 12 = (2x + 3)(x - 4) \)Denominator: \( 3x^2 - 11x - 4 = (3x + 1)(x - 4) \)

- Identify vertical asymptotes

Vertical asymptotes occur when the denominator equals zero, and the numerator does not equal zero at those points. Set the factored denominator equal to zero and solve for \( x \): \( (3x + 1)(x - 4) = 0 \)This gives two potential vertical asymptotes: \( 3x + 1 = 0 \Rightarrow x = -\frac{1}{3} \) and \( x = 4 \). Check if the numerator equals zero at these points:For \( x = -\frac{1}{3} \), \( 2(-\frac{1}{3})^2 - 5(-\frac{1}{3}) - 12 eq 0 \), so there is a vertical asymptote at \( x = -\frac{1}{3} \).For \( x = 4 \), \( 2(4)^2 - 5(4) - 12 = 0 \), thus \( x = 4 \) is not a vertical asymptote, it is a hole in the graph.

- Identify horizontal asymptotes

Horizontal asymptotes are determined by comparing the degrees of the numerator and the denominator. If the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients. For \( Q(x) \), both the numerator and denominator are quadratic (degree 2):The leading coefficient of the numerator is 2, and the leading coefficient of the denominator is 3. Thus, the horizontal asymptote is \( y = \frac{2}{3} \).

- Consider oblique asymptotes

Oblique (or slant) asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator, but in this case, the degrees are the same (both are 2), so there is no oblique asymptote for \( Q(x) \).

Key concepts

Vertical Asymptotes

Vertical asymptotes are lines that the graph of a function approaches but never touches or crosses. They occur at the values of x that make the denominator of a rational function equal to zero, provided those same x-values do not also make the numerator zero.
To find vertical asymptotes for the function \( Q(x) = \frac{2x^2 - 5x - 12}{3x^2 - 11x - 4} \), we start by factoring the denominator: \( 3x^2 - 11x - 4 = (3x + 1)(x - 4) \). Setting each factor to zero, we get the potential vertical asymptotes:

• \( 3x + 1 = 0 \rightarrow x = -\frac{1}{3} \)
• \( x - 4 = 0 \rightarrow x = 4 \)

Next, we check if the numerator is zero at these points:

• For \( x = -\frac{1}{3} \): \( 2(-\frac{1}{3})^2 - 5(-\frac{1}{3}) - 12 eq 0 \), confirming a vertical asymptote at \( x = -\frac{1}{3} \).
• For \( x = 4 \): \( 2(4)^2 - 5(4) - 12 = 0 \), which means there is a hole in the graph at \( x = 4 \).

Therefore, the only vertical asymptote of \( Q(x) \) is at \( x = -\frac{1}{3} \).

Horizontal Asymptotes

Horizontal asymptotes represent the value that a function approaches as \( x \) goes to positive or negative infinity. For rational functions, these asymptotes are found by comparing the degrees of the numerator and the denominator.
Let's determine the horizontal asymptote for \( Q(x) = \frac{2x^2 - 5x - 12}{3x^2 - 11x - 4} \):

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \( y = 0 \).
• If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is \( y = \frac{a}{b} \), where \( a \) and \( b \) are the leading coefficients of the numerator and the denominator.
• If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote (consider oblique asymptotes instead).

In our function \( Q(x) \), both the numerator and the denominator are of degree 2. The leading coefficients are 2 (numerator) and 3 (denominator), so the horizontal asymptote is calculated as: \( y = \frac{2}{3} \). Hence, the horizontal asymptote for \( Q(x) \) is at \( y = \frac{2}{3} \).

Oblique Asymptotes

Oblique asymptotes, also known as slant asymptotes, occur when a rational function's numerator degree is exactly one more than the degree of its denominator. These asymptotes are not parallel to the x or y axis.
To determine if there is an oblique asymptote for the function \( Q(x) = \frac{2x^2 - 5x - 12}{3x^2 - 11x - 4} \), observe the degrees of the numerator and the denominator. The numerator \( 2x^2 - 5x - 12 \) and the denominator \( 3x^2 - 11x - 4 \) both have a degree of 2.

• Since the degree of the numerator (2) is not one more than that of the denominator (2), there is no oblique asymptote for this function.

However, if there were an oblique asymptote, it would involve polynomial long division of the numerator by the denominator.
In summary, \( Q(x) \) does not have oblique asymptotes as their degrees do not meet the necessary criteria.