Chapter 9: Problem 23
Predict the location of any asymptotes and points of discontinuity for each function. Then, use technology to check your predictions. a) \(y=\frac{x+2}{x^{2}-4}+\frac{5}{x+2}\) b) \(y=\frac{2 x^{3}-7 x^{2}-15 x}{x^{2}-x-20}\)
Short Answer
Expert verified
For a), hole at x = -2, vertical asymptote at x = 2. For b), hole at x = 5, vertical asymptote at x = -4.
Step by step solution
01
Identify the Denominator
Examine the denominators of each term in the functions to find potential points of discontinuity and vertical asymptotes. For each one, solve for when the denominator equals zero.
02
Simplify the Expressions
Factor the denominators to identify common factors with the numerator. This helps in identifying holes (removable discontinuities) and vertical asymptotes.
03
Analysis of Asymptotes and Discontinuities - Part a
For the given function \( y = \frac{x+2}{x^2-4} + \frac{5}{x+2} \):Simplify it:1. Factor the denominator: \( x^2 - 4 = (x+2)(x-4) \)2. Rewrite the function: \( y = \frac{x+2}{(x+2)(x-2)} + \frac{5}{x+2} \) = \( \frac{x+2}{(x+2)(x-2)} + \frac{5(x-2)}{(x+2)(x-2)} = \frac{(x+2) + 5(x-2)}{(x+2)(x-2)} = \frac{6x-8}{(x+2)(x-2)} \)Potential discontinuities occur at \(x = -2\) and \(x = 2\). The term \( (x+2) \) cancels out in the original expression, so there is a hole at \(x = -2\).Vertical Asymptote at \(x = 2\).
04
Analysis of Asymptotes and Discontinuities - Part b
For the given function \( y = \frac{2x^3 - 7x^2 - 15x}{x^2 - x - 20} \):1. Factor both numerator and the denominator:\( 2x^3 - 7x^2 - 15x = x(2x^2 - 7x - 15) \)\( x^2 - x - 20 = (x-5)(x+4) \)2. Using quadratic formula, factor: \( 2x^2 - 7x - 15 = (2x+3)(x-5) \)Rewrite the function:\( y = \frac{x(2x+3)(x-5)}{(x-5)(x+4)} = \frac{x(2x+3)}{x+4} \) if \(x≠5\)Potential discontinuities occur at \(x = 5\) and \(x = -4\). The term \( (x-5) \) cancels out in the original expression, so there is a hole at \(x = 5\).Vertical asymptote at \(x = -4\).
05
Verification Using Technology
Use graphing technology to graph both functions. Check for the presence of holes at the predicted discontinuity points and vertical asymptotes at the predicted values.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
vertical asymptote
A vertical asymptote occurs where a function's value becomes undefined and shoots off towards positive or negative infinity. This typically happens when the denominator of a fraction tends to zero while the numerator does not equal zero at the same point.
For exercises like the ones provided, predicting the vertical asymptotes requires identifying the values of x that make the denominator zero and where those values do not cancel out with the numerator.
Graphing these functions or analyzing their simplified forms helps confirm the exact locations of vertical asymptotes.
For exercises like the ones provided, predicting the vertical asymptotes requires identifying the values of x that make the denominator zero and where those values do not cancel out with the numerator.
- In part a), after factoring we end up with \((x+2)(x-4)\) in the denominator. The point where the function remains undefined and results in a vertical asymptote is \(x = 2\). It does not cancel with the numerator terms.
- In part b), the denominator \(x-5)(x+4)\) suggests a vertical asymptote at \(x = -4\) since this value causes the denominator to zero without cancellation in the simplified numerator.
Graphing these functions or analyzing their simplified forms helps confirm the exact locations of vertical asymptotes.
hole in graph
A hole in a graph, known as a removable discontinuity, occurs when a factor in the numerator and the denominator cancels out. This means the function is technically undefined at one specific point, but tends to a particular value if approached from either side.
For example:
These points of cancellation indicate where the graph will have holes instead of vertical asymptotes. Graphing the function shows a small gap at these x-values, confirming the points of removable discontinuities.
For example:
- In part a), the factor \(x+2\) is present in both the numerator and denominator. When simplified, this factor cancels out, which results in a hole at \(x = -2\).
- In part b), the factor \(x-5\) cancels out in both the numerator and denominator. Thus, there is a hole at \(x = 5\).
These points of cancellation indicate where the graph will have holes instead of vertical asymptotes. Graphing the function shows a small gap at these x-values, confirming the points of removable discontinuities.
discontinuities
Discontinuities in a function are points at which the function is not defined or does not have a real value. These include vertical asymptotes as well as holes in the graph.
There are three main types of discontinuities:
Recognizing different types of discontinuities aids in graphing and understanding the overall behavior of functions. Using technology to graph these functions provides a visual confirmation of predicted discontinuities, reinforcing the problem-solving process.
There are three main types of discontinuities:
- Removable Discontinuities (Holes): Occur when a common factor in the numerator and denominator cancels out. For example, part a) has a hole at \(x = -2\) and part b) has a hole at \(x = 5\).
- Non-Removable Discontinuities: Includes vertical asymptotes which occur where the denominator is zero, and no cancelling factor exists with the numerator. As seen in part a) at \(x = 2\) and part b) at \(x = -4\).
Recognizing different types of discontinuities aids in graphing and understanding the overall behavior of functions. Using technology to graph these functions provides a visual confirmation of predicted discontinuities, reinforcing the problem-solving process.