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Chapter 11: Problem 2

Which of the following statements about the graph of \(y=\frac{x^{2}+1}{x^{2}-1}\) is not true? (A) The graph is symmetric to the \(y\) -axis. (B) There is no \(y\) -intercept. (C) The graph has one horizontal asymptote. (D) There is no \(x\) -intercept.

Short Answer

Expert verified
Statement (B) is not true.

Step by step solution

01

Determine the Symmetry of the Graph

To check if the graph is symmetric to the y-axis, substitute \(-x\) into the function: \[ y = \frac{(-x)^2 + 1}{(-x)^2 - 1} = \frac{x^2 + 1}{x^2 - 1} = y \] Since the expression remains unchanged, the graph is symmetric to the y-axis. Therefore, statement (A) is true.
02

Check for the y-intercept

To find the y-intercept, set \( x = 0 \) in the function: \[ y = \frac{0^2 + 1}{0^2 - 1} = -1 \] The graph has a y-intercept at \((0, -1)\). Therefore, statement (B) is not true.
03

Analyze the Horizontal Asymptote

For the horizontal asymptote, examine the behavior as \( x \to \infty \) or \( x \to -\infty \): As \( x \) gets large, the degree of the numerator and denominator are the same, thus, the horizontal asymptote is: \[ y = \text{leading coefficient of the numerator} / \text{leading coefficient of the denominator} = 1 \] Therefore, statement (C) is true.
04

Evaluate the x-intercept

To find the x-intercepts, set the numerator equal to 0: \[ x^2 + 1 = 0 \] Since \( x^2 + 1 eq 0 \) for any real number \( x \) (as \( x^2 = -1 \) is not possible with real numbers), there is no x-intercept. Therefore, statement (D) is true.

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Key Concepts

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

Graph Symmetry
Graphs can demonstrate various forms of symmetry, such as symmetry about the y-axis, x-axis, or the origin. In this particular exercise, we explore symmetry to the y-axis.

To determine if a graph is symmetric with respect to the y-axis, we replace every occurrence of \( x \) in the equation with \( -x \) and check if the resulting expression remains the same.

If the equation is unchanged, then the graph is y-axis symmetric.
  • For the equation \( y = \frac{x^2+1}{x^2-1} \), substituting \(-x\) gives the same function: \( y = \frac{(-x)^2+1}{(-x)^2-1} = \frac{x^2+1}{x^2-1} \).
  • This shows the graph is symmetric about the y-axis.
Understanding symmetry helps in predicting and sketching the graph, simplifying the analysis since one side reflects the other.
Horizontal Asymptote
Horizontal asymptotes are lines that a graph approaches as \( x \) moves towards positive or negative infinity. They provide insights into the behavior of functions at the extremes.

To find a horizontal asymptote, particularly for rational functions, we consider the degrees of the polynomials in the numerator and the denominator.
  • If the degrees are equal, the horizontal asymptote is found by dividing the leading coefficients of the numerator and the denominator.
  • For \( y = \frac{x^2+1}{x^2-1} \), both numerator and denominator are of degree 2.
  • The leading coefficients for both are 1. So, \( y = 1 \) is the horizontal asymptote.
This means as \( x \to \pm\infty \), the graph approaches \( y = 1 \), providing a horizontal boundary to the graph's behavior.
Y-Intercept
The y-intercept of a graph is the point where the graph intersects the y-axis. This occurs when \( x = 0 \). To find the y-intercept of a function, we substitute \( x = 0 \) into the equation and solve for \( y \).

In our example:
  • Set \( x = 0 \) in the function \( y = \frac{x^2+1}{x^2-1} \).
  • It yields \( y = \frac{0^2+1}{0^2-1} = -1 \).
  • Thus, the y-intercept is \((0, -1)\).
This point gives information about the graph's position relative to the y-axis and is often a starting point for graphing the function.
X-Intercept
The x-intercepts of a graph are points where the graph crosses the x-axis, i.e., where \( y = 0 \). For this, we solve the equation by setting the numerator equal to zero.

In rational functions, the x-intercepts occur when the numerator alone is zero because the denominator does not affect where the function equals zero, provided it's non-zero at those points.
  • For \( y = \frac{x^2+1}{x^2-1} \), we set the numerator \( x^2 + 1 = 0 \).
  • This results in \( x^2 = -1 \), which has no real solutions (only imaginary numbers).
  • Therefore, there are no real x-intercepts.
Recognizing no x-intercepts simplifies graph analysis, indicating the graph does not touch or cross the x-axis.

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