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Find the horizontal and vertical asymptotes of the graph of the function. (You need not sketch the graph.) $$ f(x)=\frac{x^{4}-x^{2}}{x(x-1)(x+2)} $$

Short Answer

Expert verified
The horizontal asymptote of the given function is \(y = 0\), and the vertical asymptotes are \(x = 0\), \(x = 1\), and \(x = -2\).

Step by step solution

01

Find the domain of the function

First, we need to find the domain of the function to know which values of x are allowed in the function. The only time the function will be undefined is when the denominator is equal to 0. So, to find the domain, we need to find the values of x for which the denominator is not equal to 0. The denominator of the function is: \(x(x-1)(x+2)\) Set the denominator equal to 0 and solve for x: \(x(x-1)(x+2) = 0\) This equation has three solutions: \(x = 0\), \(x = 1\), and \(x = -2\). So, the domain of the function is all real numbers except for these three values.
02

Find the horizontal asymptotes

To find the horizontal asymptotes, we need to analyze the behavior of the function as x approaches infinity and negative infinity. As x approaches infinity or negative infinity, the dominant terms in the numerator and the denominator will determine the behavior of the function. In this case, the dominant term in the numerator is \(x^4\) and the dominant term in the denominator is \(x^3\). We can find the limit as x approaches infinity or negative infinity by dividing the numerator and denominator by \(x^3\): \(\lim_{x \to \infty} \frac{x^4 - x^2}{x^4}\) \(\lim_{x \to \infty} \frac{1 - \frac{1}{x^2}}{x}\) As x approaches infinity, the term \(\frac{1}{x^2}\) approaches 0, so the limit becomes: \(\lim_{x \to \infty} \frac{1}{x} = 0\) Similarly, as x approaches negative infinity, the limit is also 0. Therefore, the horizontal asymptote is \(y = 0\).
03

Find the vertical asymptotes

The vertical asymptotes are found by analyzing the zeros of the denominator. We have already found the three values of x for which the denominator is equal to 0: \(x = 0\), \(x = 1\), and \(x = -2\). These are the three vertical asymptotes of the function. In conclusion, the horizontal asymptote of the given function is \(y = 0\), and the vertical asymptotes are \(x = 0\), \(x = 1\), and \(x = -2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Asymptotes
When you're working with functions and graphs, understanding horizontal asymptotes is essential. These are horizontal lines that the graph of a function approaches as the independent variable, often represented by x, goes towards positive or negative infinity. Think of them as the function's 'end behavior' on the graph — the horizontal line that the graph gets closer and closer to but never actually touches.

In our exercise example, we calculated the horizontal asymptote by examining the degrees of the polynomials in the numerator and the denominator of the function. For the function \( f(x)=\frac{x^{4}-x^{2}}{x(x-1)(x+2)} \), the highest power in the numerator is 4 (from \( x^4 \) ) and in the denominator is 3 (from \( x^3 \) ). As x grows larger and larger, those dominant terms (\( x^4 \) and \( x^3 \)) dictate the behavior of the function. This behavior can be analyzed using limits at infinity, yielding a horizontal asymptote at \( y=0 \). This indicates that no matter how far you move along the x-axis, the function's value will get nearer to 0 but will never actually be 0.
Vertical Asymptotes
Vertical asymptotes occur at specific points on the x-axis where a function's value becomes unbounded or tends to infinity. You'll find vertical asymptotes where the function is undefined, which in turn generally corresponds to zeros of the denominator in a rational function.

During our investigation of \( f(x)=\frac{x^{4}-x^{2}}{x(x-1)(x+2)} \), we identified the values that make the denominator zero: \( x=0 \), \( x=1 \), and \( x=-2 \). These values are precisely where the vertical asymptotes lie. This means as you graph the function, you'll notice that it grows infinitely large or drops to infinitely negative near these points, emphasizing that these values are not part of the function's domain.
Domain of a Function
The domain of a function constitutes all the possible input values (usually x values) which the function can accept without causing any mathematical issues, like division by zero. So, determining the domain is like drawing a boundary around what x-values are feasible.

In our example, we saw that the function \( f(x) \) has a rational form. Hence, its domain is restricted by the need to keep the denominator away from zero to avoid undefined expressions. After setting the denominator equal to zero and solving for x, we excluded \( x=0 \), \( x=1 \), and \( x=-2 \) from the domain. Consequently, the domain consists of all real numbers except these three, highlighting the importance of understanding the effects that such exclusions have on the function's graph.
Limits at Infinity
Limits at infinity provide a powerful tool to forecast the behavior of a function as x approaches extreme values (either incredibly large or incredibly small). They are a cornerstone concept in calculus and relate intimately to the topic of asymptotes.

In practical terms, when you calculate the limit of \( f(x) \) as x tends to infinity or negative infinity, you're determining where the graph of the function will ultimately settle or tend towards. Our function \( f(x)=\frac{x^{4}-x^{2}}{x(x-1)(x+2)} \) was simplified and, through limits at infinity, revealed to approach 0 as x increases or decreases without bound. This concept underscores the idea that the values of a function can stabilize to a particular number (in this case, 0) as x grows very large in either the positive or negative direction.

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