Find the Asymptotes (x^2+4x)/(3x^3-11x^2-4x)
The question asks to determine the asymptotes of the function (x^2+4x)/(3x^3-11x^2-4x). This function is a rational function (a ratio of two polynomials). Asymptotes refer to lines that the graph of the function approaches but never actually reaches. There are typically two types of asymptotes that could be relevant for this function:
Vertical asymptotes, which occur at the values of x where the denominator of the rational function is zero (unless the numerator is also zero at those points, in which case the indeterminate form might be simplified).
Horizontal or oblique asymptotes, which describe the behavior of the function as x approaches infinity or negative infinity. Horizontal asymptotes occur if the degrees of the polynomials in the numerator and denominator are the same or if the degree of the numerator is less than the denominator, while oblique asymptotes may occur when the degree of the numerator is exactly one more than the degree of the denominator.
The question is asking to identify and provide equations for these lines in relation to the given function.
$\frac{x^{2} + 4 x}{3 x^{3} - 11 x^{2} - 4 x}$
Determine the values of $x$ for which the function $f(x) = \frac{x^2 + 4x}{3x^3 - 11x^2 - 4x}$ is not defined. These are $x = -\frac{1}{3}$, $x = 0$, and $x = 4$.
Examine the limit of $f(x)$ as $x$ approaches $-\frac{1}{3}$ from both sides. The function approaches infinity from the left and negative infinity from the right, indicating a vertical asymptote at $x = -\frac{1}{3}$.
Investigate the behavior of $f(x)$ as $x$ approaches $4$. The function tends towards negative infinity from the left and infinity from the right, confirming another vertical asymptote at $x = 4$.
Compile a list of the vertical asymptotes we have found: $x = -\frac{1}{3}$ and $x = 4$.
Consider a general rational function $R(x) = \frac{ax^n}{bx^m}$, where $n$ is the degree of the numerator and $m$ is the degree of the denominator. The rules for horizontal asymptotes are as follows:
If $n < m$, the horizontal asymptote is $y = 0$.
If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$.
If $n > m$, there is no horizontal asymptote, but there could be an oblique asymptote.
Identify the degrees $n$ and $m$ for our function, where $n = 2$ and $m = 3$.
Since $n < m$, the horizontal asymptote for our function is the x-axis, or $y = 0$.
An oblique asymptote is not present because the degree of the numerator is not greater than the degree of the denominator.
Summarize the asymptotes of the function:
Asymptotes are lines that a graph approaches but does not actually reach. They can be vertical, horizontal, or oblique.
Vertical Asymptotes occur at values of $x$ where the function is undefined and the limits approach infinity or negative infinity. They can be found by setting the denominator equal to zero and solving for $x$.
Horizontal Asymptotes are found by comparing the degrees of the numerator and denominator ($n$ and $m$ respectively). If $n < m$, the horizontal asymptote is $y = 0$. If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$, where $a$ and $b$ are the leading coefficients of the numerator and denominator. If $n > m$, there is no horizontal asymptote.
Oblique Asymptotes (also called slant asymptotes) occur when the degree of the numerator is exactly one more than the degree of the denominator. They can be found by performing polynomial long division.
In the given problem, we identified the vertical asymptotes by finding the values that make the denominator zero and checking the behavior of the function around those values. We determined the horizontal asymptote by comparing the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. There is no oblique asymptote because the degree of the numerator is not one more than the degree of the denominator.