Find the Asymptotes (9x^2-4)/(3x^2+4x+1)
The question requires identifying the lines that the graph of the given rational function (9x^2-4)/(3x^2+4x+1) approaches but never actually reaches as the value of x increases or decreases without bound. These lines are called asymptotes and can be vertical, horizontal, or oblique (slant). The question is asking for a determination of the values of x and/or y where the behavior of the function can be described as approaching a specific constant value at the extremes of its domain or range, which indicates the presence of asymptotes. The process often involves finding the limits of the function as x approaches certain critical points or infinity and simplifying the expression to reveal these asymptotic behaviors.
$\frac{9 x^{2} - 4}{3 x^{2} + 4 x + 1}$
Answer
Solution:
Step 1:
Determine the values of $x$ for which the function $f(x) = \frac{9x^2 - 4}{3x^2 + 4x + 1}$ is not defined. These values are found by setting the denominator equal to zero: $3x^2 + 4x + 1 = 0$. The solutions are $x = -1$ and $x = -\frac{1}{3}$.
Step 2:
Examine the behavior of $f(x)$ as $x$ approaches $-1$. As $x$ approaches $-1$ from the left, $f(x)$ approaches infinity, and from the right, it approaches negative infinity. Thus, $x = -1$ is a vertical asymptote.
Step 3:
Analyze the behavior of $f(x)$ as $x$ approaches $-\frac{1}{3}$. As $x$ approaches $-\frac{1}{3}$ from the left, $f(x)$ approaches infinity, and from the right, it approaches negative infinity. Hence, $x = -\frac{1}{3}$ is also a vertical asymptote.
Step 4:
Compile a list of all vertical asymptotes: $x = -1$ and $x = -\frac{1}{3}$.
Step 5:
To find horizontal asymptotes for the rational function $R(x) = \frac{ax^n}{bx^m}$, apply the following rules based on the degrees $n$ and $m$ of the numerator and denominator, respectively:
If $n < m$, the horizontal asymptote is $y = 0$.
If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$.
If $n > m$, there are no horizontal asymptotes (an oblique asymptote may exist).
Step 6:
Identify the degrees $n$ and $m$ of the numerator and denominator. In this case, $n = 2$ and $m = 2$.
Step 7:
Since $n$ equals $m$, the horizontal asymptote is determined by the ratio of the leading coefficients $a$ and $b$. Here, $a = 9$ and $b = 3$, so the horizontal asymptote is $y = \frac{a}{b} = 3$.
Step 8:
Conclude that there are no oblique asymptotes since the degree of the numerator is not greater than the degree of the denominator.
Step 9:
Summarize the complete set of asymptotes for the function:
- Vertical Asymptotes: $x = -1$, $x = -\frac{1}{3}$
- Horizontal Asymptote: $y = 3$
- No Oblique Asymptotes
Knowledge Notes:
The concept of asymptotes is important in understanding the behavior of functions as the input values become very large in magnitude. There are three types of asymptotes:
Vertical Asymptotes: These occur at values of $x$ where the function goes to infinity or negative infinity. They are found by setting the denominator of a rational function to zero and solving for $x$.
Horizontal Asymptotes: These describe the behavior of a function as $x$ goes to infinity or negative infinity. The rules for finding horizontal asymptotes depend on the degrees of the polynomial in the numerator ($n$) and the denominator ($m$):
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, respectively.
If $n > m$, there are no horizontal asymptotes.
Oblique Asymptotes: These occur when the degree of the numerator is exactly one more than the degree of the denominator. The oblique asymptote can be found by performing polynomial long division.
In the given problem, we found the vertical asymptotes by solving for the values that make the denominator zero. We then determined the horizontal asymptote by comparing the degrees of the numerator and denominator and using the leading coefficients. Since the degrees are equal, the horizontal asymptote is $y = \frac{a}{b}$. There are no oblique asymptotes in this case because the degree of the numerator is not greater than the degree of the denominator.
