Find the Asymptotes (7x)/(x^3-5x^2)
The question is asking to determine the asymptotes of the function \( f(x) = \frac{7x}{x^3 - 5x^2} \). An asymptote is a line that the graph of a function approaches but never touches as the independent variable (x in this case) approaches infinity or a certain value. There are different kinds of asymptotes - vertical, horizontal, and oblique (slant). Vertical asymptotes occur at values of x where the function becomes undefined, typically where the denominator is zero. Horizontal asymptotes are determined by the behavior of the function as x approaches infinity or negative infinity, indicating the constant value the function is approaching. Oblique asymptotes occur when the function approaches a line that is not horizontal or vertical, usually when the degree of the numerator is one higher than the degree of the denominator. In this case, the question may involve finding all of these types of asymptotes depending on the behavior of the function.
$\frac{7 x}{x^{3} - 5 x^{2}}$
Answer
Solution:
Step 1:
Identify the values of $x$ that cause the function $\frac{7x}{x^3 - 5x^2}$ to be undefined, which are the roots of the denominator. These are $x = 0$ and $x = 5$.
Step 2:
Observe the behavior of the function as $x$ approaches $0$. As $x \to 0^-$, $\frac{7x}{x^3 - 5x^2} \to \infty$ and as $x \to 0^+$, $\frac{7x}{x^3 - 5x^2} \to -\infty$. Hence, $x = 0$ is a vertical asymptote.
Step 3:
Examine the behavior of the function as $x$ approaches $5$. As $x \to 5^-$, $\frac{7x}{x^3 - 5x^2} \to -\infty$ and as $x \to 5^+$, $\frac{7x}{x^3 - 5x^2} \to \infty$. Therefore, $x = 5$ is another vertical asymptote.
Step 4:
Compile a list of all vertical asymptotes found: $x = 0$ and $x = 5$.
Step 5:
Consider the general form of a rational function $R(x) = \frac{ax^n}{bx^m}$ to determine horizontal asymptotes based on the degrees of the numerator ($n$) and denominator ($m$):
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 possibly an oblique asymptote.
Step 6:
Determine the degrees $n$ and $m$ for the given function: $n = 1$ and $m = 3$.
Step 7:
Since $n < m$, conclude that the horizontal asymptote is the x-axis, $y = 0$.
Step 8:
Confirm that there is no oblique asymptote, as the degree of the numerator is less than the degree of the denominator.
Step 9:
Summarize the asymptotes of the function:
Vertical Asymptotes: $x = 0$ and $x = 5$ Horizontal Asymptote: $y = 0$ No Oblique Asymptotes
Knowledge Notes:
Asymptotes are lines that a graph of a function approaches as the independent variable (usually $x$) goes to infinity or negative infinity. There are three types of asymptotes: vertical, horizontal, and oblique.
Vertical Asymptotes occur at values of $x$ where the function becomes undefined, typically where the denominator of a rational function is zero. The function's value approaches infinity or negative infinity near these points.
Horizontal Asymptotes are found by comparing the degrees of the numerator and the denominator in a rational function. If the degree of the numerator ($n$) is less than the degree of the denominator ($m$), 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 is no horizontal asymptote.
Oblique Asymptotes (or slant asymptotes) may occur when the degree of the numerator is exactly one more than the degree of the denominator. In such cases, polynomial long division can be used to find the equation of the oblique asymptote.
In the given problem, we use these principles to find the vertical and horizontal asymptotes of the function $\frac{7x}{x^3 - 5x^2}$. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at $y = 0$, and no oblique asymptote exists.
