Find the Asymptotes f(x)=(2x)/(9x^2+1)
The question requires you to identify the asymptotes of the function f(x) = (2x)/(9x^2 + 1). An asymptote is a line that the graph of a function approaches but never touches as the values of x increase or decrease without bound. There are different types of asymptotes, including horizontal, vertical, and oblique (slant) asymptotes.
For the given rational function, you would typically look for:
Vertical asymptotes by determining the values of x that would make the denominator equal to zero, if any.
Horizontal asymptotes by comparing the degrees of the polynomial in the numerator and the denominator as x approaches infinity or negative infinity.
Oblique (slant) asymptotes which may occur if the degree of the polynomial in the numerator is exactly one degree higher than the denominator.
The question is inviting you to conduct this analysis for the given function and report the equations of any asymptotes found.
$f \left(\right. x \left.\right) = \frac{2 x}{9 x^{2} + 1}$
Answer
Solution:
Step 1:
Determine the values for which $\frac{2x}{9x^2 + 1}$ does not exist. The function is defined for all real numbers since the denominator never equals zero.
Step 2:
Identify any vertical asymptotes by finding points of infinite discontinuity. There are no vertical asymptotes for this function.
Step 3:
Examine the 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.
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; instead, there may be an oblique asymptote.
Step 4:
Calculate the values of $n$ and $m$.
- $n = 1$
- $m = 2$
Step 5:
Since $n < m$, the horizontal asymptote is the x-axis, given by $y = 0$.
Step 6:
There is no slant or oblique asymptote since the degree of the numerator is less than the degree of the denominator.
Step 7:
Summarize the asymptotes of the function:
- No Vertical Asymptotes
- Horizontal Asymptote: $y = 0$
- No Oblique Asymptotes
Knowledge Notes:
To find the asymptotes of a rational function, we must consider both the vertical and horizontal (or oblique) asymptotes.
Vertical Asymptotes: These occur at values of $x$ that cause the denominator of the rational function to be zero, provided that the numerator is not also zero at those points. They represent values where the function approaches infinity.
Horizontal Asymptotes: These are found by comparing the degrees of the numerator and denominator ($n$ and $m$, respectively). If the degree of the numerator is less than the degree of the denominator ($n < m$), the horizontal asymptote is $y = 0$. If the degrees are equal ($n = m$), the horizontal asymptote is $y = \frac{a}{b}$, where $a$ and $b$ are the leading coefficients of the numerator and denominator, respectively. If the numerator's degree is greater than the denominator's ($n > m$), there is no horizontal asymptote.
Oblique Asymptotes: These occur when the degree of the numerator is exactly one more than the degree of the denominator ($n = m + 1$). To find the equation of an oblique asymptote, you would typically perform polynomial long division.
In the given problem, the function $\frac{2x}{9x^2 + 1}$ has no vertical asymptotes because the denominator cannot be zero. The horizontal asymptote is $y = 0$ because the degree of the numerator ($n = 1$) is less than the degree of the denominator ($m = 2$). There are no oblique asymptotes because the degree of the numerator is not greater than the degree of the denominator.
