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Chapter 2: Problem 34

(a) Find the vertical asymptotes of the function $$y=\frac{x^{2}+1}{3 x-2 x^{2}}$$ (b) Confirm your answer to part (a) by graphing the function.

Short Answer

Expert verified
Vertical asymptotes are at \( x = 0 \) and \( x = \frac{3}{2} \). Graphing confirms these asymptotes.

Step by step solution

01

Determine the Criteria for Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function equals zero, provided that the numerator is not also zero at those values. For the function \( y = \frac{x^2 + 1}{3x - 2x^2} \), set the denominator equal to zero: \( 3x - 2x^2 = 0 \).
02

Solve the Denominator Equation

To solve for \( x \) where the denominator is zero, rearrange it as \( -2x^2 + 3x = 0 \). Factor out \( x \) to get \( x(3 - 2x) = 0 \). This gives the solutions \( x = 0 \) and \( x = \frac{3}{2} \).
03

Check the Numerator at These Points

Next, check the numerator at \( x = 0 \) and \( x = \frac{3}{2} \). The numerator is \( x^2 + 1 \), which equals 1 at \( x = 0 \) and \( \left(\frac{3}{2}\right)^2 + 1 = \frac{9}{4} + 1 = \frac{13}{4} \) at \( x = \frac{3}{2} \). Since neither evaluation yields zero, both points are vertical asymptotes for the function.
04

Graph the Function to Confirm

Graph the function \( y = \frac{x^2 + 1}{3x - 2x^2} \). Check that as \( x \) approaches the vertical asymptotes \( x = 0 \) and \( x = \frac{3}{2} \), the value of \( y \) approaches infinity or negative infinity, confirming these are indeed the vertical asymptotes.

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Key Concepts

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

Rational Functions
Rational functions are a type of function represented by the ratio of two polynomials. They take the form \( y = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are both polynomials and \( Q(x) eq 0 \). It is important that the denominator is not zero as division by zero is undefined.

In the given function \( y = \frac{x^2 + 1}{3x - 2x^2} \), \( x^2 + 1 \) is the numerator and \( 3x - 2x^2 \) is the denominator. Rational functions can have horizontal, vertical, or slant asymptotes, which help describe their graphs.
  • Vertical asymptotes occur where the denominator is zero and the numerator is not.
  • Horizontal asymptotes are found based on the degrees of the numerator and denominator.
Understanding these properties helps in graphing the function and predicting its behavior.
Denominator
The denominator of a rational function is a crucial part since it defines where the function is undefined. When the denominator equals zero, the function cannot have a finite value. This specific point dictates the existence of vertical asymptotes.

In our example with the denominator \( 3x - 2x^2 \), by setting it to zero, we find the values \( x = 0 \) and \( x = \frac{3}{2} \), indicating possible vertical asymptotes. The process to find these values involves solving the quadratic equation for these points.
  • Factor the equation \(-2x^2 + 3x = 0\).
  • Extract common terms or use the quadratic formula if necessary.
  • Verify if the numerator is zero at these points to confirm the asymptotes.
By ensuring the numerator is not zero at these points, we confirm the location of vertical asymptotes.
Numerator
The numerator in a rational function influences whether a vertical asymptote truly exists at a point where the denominator is zero.

For our function, \( x^2 + 1 \) is the numerator. Checking this at the points \( x = 0 \) and \( x = \frac{3}{2} \) helps ensure there is no simplification that removes the asymptote.
  • Evaluate \( x^2 + 1 \) at \( x = 0 \): the result is 1.
  • Evaluate at \( x = \frac{3}{2} \): the result is \( \frac{13}{4} \).
Neither of these evaluations results in zero, confirming the presence of vertical asymptotes at both points. This step is critical to confirming the asymptotal behavior rather than having potential holes in the graph.
Graphing Functions
Graphing functions like rational functions allows you to visually confirm properties like vertical asymptotes.

To graph \( y = \frac{x^2 + 1}{3x - 2x^2} \), observe how \( y \) behaves as \( x \) approaches the vertical asymptotes at \( x = 0 \) and \( x = \frac{3}{2} \).
  • As \( x \to 0^+ \), \( y \to +\infty \) and as \( x \to 0^- \), \( y \to -\infty \).
  • Similarly, as \( x \to \frac{3}{2}^+ \), \( y \to -\infty \), and \( x \to \frac{3}{2}^- \), \( y \to +\infty \).
This behavior confirms the asymptotes since the graph shoots upwards or downwards indefinitely. Graphing helps solidify understanding by visually connecting the algebraic function to its geometric representation on a coordinate plane.

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Most popular questions from this chapter

Show by means of an example that \(\lim _{x \rightarrow a}[f(x) g(x)]\) may exist even though neither \(\lim _{x \rightarrow a} f(x)\) nor \(\lim _{x \rightarrow a} g(x)\) exists.

Determine the infinite limit. $$\lim _{x \rightarrow-3^{-}} \frac{x+2}{x+3}$$

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \(g(\mathrm{t})=\frac{1}{\sqrt{\mathrm{t}}}\)

Is there a number a such that $$\lim _{x \rightarrow-2} \frac{3 x^{2}+a x+a+3}{x^{2}+x-2}$$ exists? If so, find the value of a and the value of the limit.

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \(f(x)=\frac{1}{2} x-\frac{1}{3}\)
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