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Chapter 2: Problem 31

Find the vertical and horizontal asymptotes for the graph of \(f\). $$ f(x)=\frac{1}{x^{3}+x^{2}-6 x} $$

Short Answer

Expert verified
Vertical asymptotes at x = 0, x = -3, x = 2; horizontal asymptote at y = 0.

Step by step solution

01

Factor the Denominator

The first step to find the asymptotes is to factor the denominator of the function. The function is given by \[ f(x) = \frac{1}{x^3 + x^2 - 6x}. \]Factor the expression \( x^3 + x^2 - 6x \) by taking out the greatest common factor (GCF), which is \( x \).This gives:\[ x(x^2 + x - 6). \] Now factor \( x^2 + x - 6 \):- Find two numbers that multiply to \(-6\) and add to \(1\).- The numbers \(3\) and \(-2\) satisfy these conditions.- So, \( x^2 + x - 6 = (x + 3)(x - 2). \)Thus, the fully factored form of the denominator is:\[ x(x + 3)(x - 2).\]
02

Find Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero but the numerator is not zero.Since the numerator is 1 (which is never zero), vertical asymptotes are located where the denominator \(x(x + 3)(x - 2) = 0 \). Solve this by setting each factor of the denominator equal to zero:- \(x = 0\)- \(x + 3 = 0 \Rightarrow x = -3\)- \(x - 2 = 0 \Rightarrow x = 2\)Thus, the vertical asymptotes are at \(x = 0\), \(x = -3\), and \(x = 2\).
03

Find Horizontal Asymptote

Horizontal asymptotes are determined by comparing the degree of the polynomial in the numerator and the denominator.- The numerator of \(f(x)\) is constant (degree 0),- The denominator is degree 3 (\(x^3 + x^2 - 6x\)).Since the degree of the denominator (3) is greater than the degree of the numerator (0), the horizontal asymptote is:\[ y = 0.\]

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Key Concepts

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

Vertical Asymptotes
Vertical asymptotes occur in a rational function where the denominator is zero and the numerator is not zero. For the function \( f(x) = \frac{1}{x^3 + x^2 - 6x} \), we can determine the vertical asymptotes by setting the denominator equal to zero. First, factor the polynomial in the denominator as \( x(x + 3)(x - 2) \). This gives us three factors that can each independently cause the denominator to be zero.
  • Set \( x = 0 \).
  • Set \( x + 3 = 0 \Rightarrow x = -3 \).
  • Set \( x - 2 = 0 \Rightarrow x = 2 \).
These calculations show that there are vertical asymptotes at \( x = 0 \), \( x = -3 \), and \( x = 2 \). In a graph, the curve would sharply approach these lines but never actually cross them.
Horizontal Asymptotes
Horizontal asymptotes in a rational function are determined by the relative degrees of the polynomials in the numerator and denominator. A horizontal asymptote represents a value that the function approaches as \( x \) becomes very large (positive or negative). In the function \( f(x) = \frac{1}{x^3 + x^2 - 6x} \):
  • The degree of the numerator is 0 (constant of 1).
  • The degree of the denominator is 3 (given by \( x^3 + x^2 - 6x \)).
Because the degree of the denominator is greater than that of the numerator, the horizontal asymptote is at \( y = 0 \). This means that as \( x \) goes to infinity or negative infinity, \( f(x) \) will approach \( y = 0 \).
Factoring Polynomials
Factoring polynomials is a crucial step for simplifying expressions and finding asymptotes. In the given problem, we start with the polynomial \( x^3 + x^2 - 6x \). The first step in factoring is to take out the greatest common factor, which in this case is \( x \), resulting in \( x(x^2 + x - 6) \).To further factor \( x^2 + x - 6 \), look for two numbers that multiply to \(-6\) and add up to 1. These numbers are \( 3 \) and \(-2 \). Thus, the quadratic is factored to \((x + 3)(x - 2)\). Therefore, the entire expression is factored as \( x(x + 3)(x - 2) \). Factoring helps identify roots of the polynomial which are essential in locating vertical asymptotes.
Degree of Polynomials
The degree of a polynomial is the highest degree of its respective terms when the polynomial is expressed in its standard form (expanded without any factored terms). In the function鈥檚 denominator \( x^3 + x^2 - 6x \), the term \( x^3 \) has the highest power, giving it a degree of 3.
  • In rational functions, the degree of the polynomial can affect both horizontal asymptotes and other behaviors of the function at infinity.
  • If the degree of the numerator is less than that of the denominator, the horizontal asymptote will be \( y = 0 \).
This understanding is fundamental, as it tells us how the function behaves for very large or very small values of \( x \), providing insight into its graph's end behavior.

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Most popular questions from this chapter

Sketch the graph of \(f\) and find each limit, if it exists: (a) \(\lim _{x \rightarrow 1} f(x)\) (b) \(\lim _{x \rightarrow 1^{+}} f(x)\) (c) \(\lim _{x \rightarrow 1} f(x)\) $$ f(x)=\left\\{\begin{array}{ll} 3 x-1 & \text { if } x \leq 1 \\ 3-x & \text { if } x>1 \end{array}\right. $$

Use the sandwich theorem to verify the limit. $$ \lim _{x \rightarrow 0}\left(x^{2}+1\right)=1 \quad\left(\text { Hint: } \text { Use } \lim _{x \rightarrow 0}(|x|+1)=1 .\right) $$

Salt water of concentration 0.1 pound of salt per gallon flows into a large tank that initially contains 50 gallons of pure water. (a) If the flow rate of salt water into the tank is 5 gallons per minute, find the volume \(V(t)\) of water and the amount \(A(t)\) of salt in the tank after \(t\) minutes. (b) Find a formula for the salt concentration \(c(t)\) (in Ib/gal) after \(t\) minutes. (c) What happens to \(c(t)\) over a long period of time?

Sketch the graph of \(f\) and find each limit, if it exists: (a) \(\lim _{x \rightarrow 1} f(x)\) (b) \(\lim _{x \rightarrow 1^{+}} f(x)\) (c) \(\lim _{x \rightarrow 1} f(x)\) $$ f(x)=\left\\{\begin{array}{ll} x^{2}-1 & \text { if } x<1 \\ 4-x & \text { if } x \geq 1 \end{array}\right. $$

Prove that the equation \(x^{5}-3 x^{4}-2 x^{3}-x+1=0\) has a solution between 0 and 1 .
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