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Chapter 9: Problem 4

Graph each function using technology and identify any asymptotes and intercepts. a) \(y=\frac{2 x+1}{x-4}\) b) \(y=\frac{3 x-2}{x+1}\) c) \(y=\frac{-4 x+3}{x+2}\) d) \(y=\frac{2-6 x}{x-5}\)

Short Answer

Expert verified
For each function: (a) Asymptotes: x=4, y=2; Intercepts: x=-1/2, y=-1/4(b) Asymptotes: x=-1, y=3; Intercepts: x=2/3, y=-2(c) Asymptotes: x=-2, y=-4; Intercepts: x=3/4, y=3/2(d) Asymptotes: x=5, y=-6; Intercepts: x=1/3, y=-2/5

Step by step solution

01

Graph the Function

Use graphing technology (a graphing calculator or software) to plot each function. Observe the shape and features of each graph carefully.
02

Determine Vertical Asymptotes

Vertical asymptotes occur where the denominator of the function equals zero, causing the function to be undefined. Set the denominator equal to zero and solve for x.
03

Determine Horizontal Asymptotes

For rational functions, compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients.
04

Find the Intercepts

To find x-intercepts, set the numerator equal to zero and solve for x. For y-intercepts, evaluate the function at x=0.
05

Analyze Each Function Separately

Let's proceed to analyze each given function.
06

Function (a) Analysis

For the function \(y=\frac{2x+1}{x-4}\): 1. Vertical asymptote: Set x-4=0, thus x=4. 2. Horizontal asymptote: Degrees are equal, so it is y=\frac{2}{1}=2. 3. X-intercept: Set 2x+1=0, thus x=-\frac{1}{2}. 4. Y-intercept: Evaluate at x=0, thus y=\frac{2(0)+1}{0-4}=-\frac{1}{4}.
07

Function (b) Analysis

For the function \(y=\frac{3x-2}{x+1}\): 1. Vertical asymptote: Set x+1=0, thus x=-1. 2. Horizontal asymptote: Degrees are equal, so it is y=\frac{3}{1}=3. 3. X-intercept: Set 3x-2=0, thus x=\frac{2}{3}. 4. Y-intercept: Evaluate at x=0, thus y=\frac{3(0)-2}{0+1}=-2.
08

Function (c) Analysis

For the function \(y=\frac{-4x+3}{x+2}\): 1. Vertical asymptote: Set x+2=0, thus x=-2. 2. Horizontal asymptote: Degrees are equal, so it is y=\frac{-4}{1}=-4. 3. X-intercept: Set -4x+3=0, thus x=\frac{3}{4}. 4. Y-intercept: Evaluate at x=0, thus y=\frac{-4(0)+3}{0+2}=\frac{3}{2}.
09

Function (d) Analysis

For the function \(y=\frac{2-6x}{x-5}\): 1. Vertical asymptote: Set x-5=0, thus x=5. 2. Horizontal asymptote: Degrees are equal, so it is y=\frac{-6}{1}=-6. 3. X-intercept: Set 2-6x=0, thus x=\frac{1}{3}. 4. Y-intercept: Evaluate at x=0, thus y=\frac{2-6(0)}{0-5}=-\frac{2}{5}.

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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 where the function becomes undefined, which happens when the denominator is zero. For a rational function like \( y = \frac{p(x)}{q(x)} \), setting \( q(x) = 0 \) and solving for \( x \) will give you the vertical asymptotes. For example, for the function \( y = \frac{2x+1}{x-4} \), setting \( x-4 = 0 \) gives \( x = 4 \). Thus, \( x = 4 \) is a vertical asymptote. Practically, vertical asymptotes are lines where the graph shoots off to infinity or negative infinity.
Horizontal Asymptotes
Horizontal asymptotes show the behavior of the function as \( x \) approaches infinity or negative infinity. They're determined by comparing the degrees of the numerator and denominator of the rational function. For \( y = \frac{2x + 1}{x - 4} \), both the numerator and denominator have a degree of 1. Here, the horizontal asymptote is the ratio of the leading coefficients, \( y = \frac{2}{1} = 2 \). If the numerator's degree is less than the denominator's, the horizontal asymptote is \( y = 0 \).
X-Intercepts
X-intercepts are points where the graph crosses the x-axis, which occur when \( y = 0 \). To find them, set the numerator equal to zero and solve for \( x \). For \( y = \frac{3x-2}{x+1} \), setting \( 3x - 2 = 0 \) and solving gives \( x = \frac{2}{3} \). This means the graph crosses the x-axis at \( x = \frac{2}{3} \). Understanding x-intercepts is crucial as they give insights where the function value is zero.
Y-Intercepts
Y-intercepts are points where the graph crosses the y-axis, which occur when \( x = 0 \). To find them, evaluate the function at \( x = 0 \). For \( y = \frac{-4x+3}{x+2} \), substituting \( x = 0 \) gives \( y = \frac{3}{2} \). Therefore, the graph crosses the y-axis at \( y = \frac{3}{2} \). Y-intercepts illustrate the initial value or starting point of the function where x is zero.
Rational Function Analysis
Analyzing rational functions involves understanding their properties and behaviors, such as vertical and horizontal asymptotes, x-intercepts, and y-intercepts. For example, let's examine \( y = \frac{2-6x}{x-5} \):
  • Vertical Asymptote: Set \( x-5=0 \), so \( x=5 \).
  • Horizontal Asymptote: Degrees are equal, take ratio of leading coefficients so \( y= -6 \).
  • X-intercept: Set \(2 - 6x = 0 \), solving for \( x = \frac{1}{3} \).
  • Y-intercept: Substitute \( x = 0 \), result is \( y = -\frac{2}{5} \).
This structured analysis helps breaking down complex functions into understandable parts.

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Most popular questions from this chapter

In hockey, a player's shooting percentage is given by dividing the player's total goals scored by the player's total shots taken on goal. So far this season, Rachel has taken 28 shots on net but scored only 2 goals. She has set a target of achieving a \(30 \%\) shooting percentage this season. a) Write a function for Rachel's shooting percentage if \(x\) represents the number of shots she takes from now on and she scores on half of them. b) How many more shots will it take for her to bring her shooting percentage up to her target?

The intensity, \(I\), of light, in watts per square metre \(\left(\mathrm{W} / \mathrm{m}^{2}\right),\) at a distance, \(d,\) in metres, from the point source is given by the formula \(I=\frac{P}{4 \pi d^{2}},\) where \(P\) is the average power of the source, in watts. How far away from a 500 -W light source is intensity \(5 \mathrm{W} / \mathrm{m}^{2} ?\)

A truck leaves Regina and drives eastbound. Due to road construction, the truck takes \(2 \mathrm{h}\) to travel the first \(80 \mathrm{km} .\) Once it leaves the construction zone, the truck travels at \(100 \mathrm{km} / \mathrm{h}\) for the rest of the trip. a) Let \(v\) represent the average speed, in kilometres per hour, over the entire trip and \(t\) represent the time, in hours, since leaving the construction zone. Write an equation for \(v\) as a function of \(t\) b) Graph the function for an appropriate domain. c) What are the equations of the asymptotes in this situation? Do they have meaning in this situation? Explain. d) How long will the truck have to drive before its average speed is \(80 \mathrm{km} / \mathrm{h} ?\) e) Suppose your job is to develop GPS technology. How could you use these types of calculations to help travellers save fuel?

Use a graphical method to solve each equation. Then, use another method to verify your solution. a) \(\frac{8}{x}-4=x+3\) b) \(2 x=\frac{10 x}{2 x-1}\) c) \(\frac{3 x^{2}+4 x-15}{x+3}=2 x-1\) d) \(\frac{3}{5 x-7}+x=1+\frac{x^{2}-4 x}{7-5 x}\)

\- Mira uses algebra to rewrite the function \(y=\frac{2-3 x}{x-7}\) in an equivalent form that she can graph by hand. \(y=\frac{2-3 x}{x-7}\) \(y=\frac{-3 x+2}{x-7}\) \(y=\frac{-3 x-21+21+2}{x-7}\) \(y=\frac{-3(x-7)+23}{x-7}\) \(y=\frac{-3(x-7)}{x-7}+\frac{23}{x-7}\) \(y=-3+\frac{23}{x-7}\) \(y=\frac{23}{x-7}-3\) a) Identify and correct any errors in Mira's work. b) How might Mira have discovered that she had made an error without using technology? How might she have done so with technology?
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