Chapter 4: Problem 30
Sketch the graph of each rational function. Specify the intercepts and the asymptotes. (a) \(y=\left(4 x^{2}+1\right) /\left(x^{2}-1\right)\) (b) \(y=\left(4 x^{2}+1\right) /\left(x^{2}+1\right)\)
Short Answer
Expert verified
(a) No x-intercepts, y-intercept at (0,-1), asymptotes: x=±1, y=4.
(b) No x-intercepts, y-intercept at (0,1), asymptote: y=4.
Step by step solution
01
Analyze Intercepts and Asymptotes for (a)
For the function \( y = \frac{4x^2 + 1}{x^2 - 1} \), first determine the **x-intercepts** by setting the numerator equal to zero: \( 4x^2 + 1 = 0 \). This equation has no real solutions since \(4x^2 + 1\) is always positive. For the **y-intercept**, substitute \(x = 0\) into the equation: \( y = \frac{4(0)^2 + 1}{(0)^2 - 1} = -1 \). Determine the **vertical asymptotes** by setting the denominator to zero: \(x^2 - 1 = 0\), which gives \( x = \pm 1 \). The **horizontal asymptote** is found as the degrees of the numerator and denominator are the same. The leading coefficients ratio is \(\frac{4}{1} = 4\), so \( y = 4 \).
02
Sketch the Graph for (a)
Plot these findings on a coordinate plane: - The vertical asymptotes at \( x = 1 \) and \( x = -1 \).- The horizontal asymptote at \( y = 4 \).- The y-intercept at \( (0, -1) \).Since there are no x-intercepts, observe how the function behaves near the asymptotes and y-intercept to complete the graph.
03
Analyze Intercepts and Asymptotes for (b)
For the function \( y = \frac{4x^2 + 1}{x^2 + 1} \), set the numerator equal to zero to find **x-intercepts**: \( 4x^2 + 1 = 0 \). This has no real solutions. Find the **y-intercept** by substituting \(x = 0\): \( y = \frac{4(0)^2 + 1}{(0)^2 + 1} = 1 \). Determine **vertical asymptotes** by setting the denominator to zero: \(x^2 + 1 = 0\), which has no real solutions, so there are no vertical asymptotes. The **horizontal asymptote** is \( y = 4 \), derived from the ratio of the leading coefficients since the degree of the numerator and denominator is the same.
04
Sketch the Graph for (b)
Plot the findings on a coordinate plane: - The horizontal asymptote at \( y = 4 \).- The y-intercept at \( (0, 1) \).Since there are no x-intercepts or vertical asymptotes, the graph is a smooth curve approaching \( y = 4 \) as \( x \to \pm \infty \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graph Sketching of Rational Functions
Graph sketching involves understanding the overall shape of a function on a coordinate plane. Rational functions can be a bit tricky because of their asymptotes and intercepts. Here are some steps to help you sketch these graphs:
- Find the **intercepts** by setting the numerator equal to zero for x-intercepts and sub in zero for y-intercepts.
- Identify **vertical asymptotes** by setting the denominator to zero.
- Determine **horizontal or slant asymptotes** based on the degree of the numerator and denominator.
- Plot these key features on the graph and understand the behavior of the function around these points.
Understanding Intercepts
Intercepts are points where the graph of a function crosses the axes. In rational functions, they are essential in setting up your graph's initial structure.
X-Intercepts
For rational functions, x-intercepts are found by setting the numerator equal to zero. However, not all numerators have real solutions, as seen in our examples. This lack of x-intercepts affects how the graph is structured since it never touches the x-axis.Y-Intercepts
These are easier to find. Simply set x to zero in the function. For example, for the function \( y = \frac{4x^2 + 1}{x^2-1} \), substituting x gives a y-intercept of -1. Likewise, for \( y = \frac{4x^2 + 1}{x^2+1} \), it yields a y-intercept of 1. Here, the graph will cross the y-axis at these points.Importance of Asymptotes
Asymptotes provide insights into the infinite behavior of rational functions. Understanding them helps in comprehending how the graph behaves over larger values of x.