Find the Asymptotes f(x)=(x(x-1))/(x^3+25x)
The given problem is asking for the determination of asymptotes of the function \( f(x) = \frac{x(x - 1)}{x^3 + 25x} \). Asymptotes are lines that the graph of the function approaches as the independent variable (in this case, x) heads towards infinity or negative infinity or some critical value where the function is undefined. There are different types of asymptotes:
Vertical Asymptotes: These occur at points where the function is undefined due to a zero in the denominator that does not cancel out with a zero in the numerator.
Horizontal Asymptotes: These indicate the behavior of the function as x approaches positive or negative infinity, showing the value that f(x) is getting closer to in the horizontal direction.
Oblique (Slant) Asymptotes: If the degree of the polynomial in the numerator is exactly one more than the degree of the polynomial in the denominator, the function may have an oblique asymptote, which is a straight line that the graph approaches as x becomes very large or very small.
The problem requires analyzing the function to find all types of asymptotes that apply to this particular rational function.
$f \left(\right. x \left.\right) = \frac{x \left(\right. x - 1 \left.\right)}{x^{3} + 25 x}$
Answer
Solution:
Step 1:
Determine the points at which the function $\frac{x(x - 1)}{x^3 + 25x}$ does not exist. This occurs when $x = 0$.
Step 2:
Identify the vertical asymptotes by locating the values that cause the function to approach infinity. There are no vertical asymptotes in this case.
Step 3:
Examine the degrees of the numerator and denominator in the function $R(x) = \frac{ax^n}{bx^m}$. The horizontal asymptote depends on the relationship between $n$ and $m$:
If $n < m$, the horizontal asymptote is $y = 0$.
If $n = m$, the horizontal asymptote is $y = \frac{a}{b}$.
If $n > m$, there are no horizontal asymptotes; instead, there may be an oblique asymptote.
Step 4:
Calculate the degrees of the numerator and denominator. Here, $n = 2$ and $m = 3$.
Step 5:
Given that $n < m$, the horizontal asymptote is the x-axis, which is $y = 0$.
Step 6:
An oblique asymptote is not present because the degree of the numerator is not greater than the degree of the denominator.
Step 7:
Compile the complete list of asymptotes for the function:
- No Vertical Asymptotes
- Horizontal Asymptote: $y = 0$
- No Oblique Asymptotes
Knowledge Notes:
The concept of asymptotes is related to the behavior of a graph as the inputs approach certain values or infinity. Asymptotes can be vertical, horizontal, or oblique:
Vertical Asymptotes: These occur at values of $x$ where the function approaches infinity. To find them, look for values that make the denominator zero but not the numerator.
Horizontal Asymptotes: These are found by comparing the degrees of the numerator and the 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, 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. They can be found by performing polynomial long division of the numerator by the denominator.
In the given problem, the function is $\frac{x(x - 1)}{x^3 + 25x}$. To find where the function is undefined, we set the denominator equal to zero and solve for $x$. However, in this case, the $x$ terms cancel out, leaving no vertical asymptotes.
For horizontal asymptotes, we compare the degrees of the numerator and denominator. Since the degree of the numerator ($n = 2$) is less than the degree of the denominator ($m = 3$), the horizontal asymptote is $y = 0$.
There are no oblique asymptotes because the degree of the numerator is not one more than the degree of the denominator.
