Difference between revisions of "005 Sample Final A, Question 20"

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(Created page with "'''Question ''' Consider the following rational function, <center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br>      a. What is the domain of f?...")
 
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<center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br>
 
<center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br>
  
&nbsp;&nbsp;&nbsp;&nbsp; a. What is the domain of f? <br>
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:: a. What is the domain of f? <br>
&nbsp;&nbsp;&nbsp;&nbsp; b. What are the x and y-intercepts of f? <br>
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:: b. What are the x and y-intercepts of f? <br>
&nbsp;&nbsp;&nbsp;&nbsp; c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes? <br>
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:: c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes? <br>
&nbsp;&nbsp;&nbsp;&nbsp; d. Graph f(x). Make sure to include the information you found above.
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:: d. Graph f(x). Make sure to include the information you found above.
  
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answers
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!Foundations: &nbsp;
 
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|-
|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
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|1) What are the asymptotes and zeros?
 
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|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
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|Answer:
 
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|c) False. <math>y = x^2</math> does not have an inverse.
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|1) The vertical asymptote corresponds to zeros of the denominator. So there is a vertical asymptote at x = -1. The zero is  at (1, 0). The horizontal asymptote is the ratio of leading coefficients. So the horizontal asymptote is <math>y = \frac{1}{2}</math>
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|}
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Solution:
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 1: &nbsp;
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|-
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|We start by finding the asymptotes. The vertical asymptote corresponds to zeros of the denominator. So the vertical asymptote is at x = -1. They horizontal asymptote is determined by degree of the numerator and degree of the denominator. Since both of those values are 1, the horizontal asymptote is the ratio of leading coefficients. This means the horizontal asymptote is <math>y = \frac{1}{2}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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|-
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|Now we observe that the zero is at (1, 0), and proceed by looking at the intervals created by removing x = -1 and x = 1. This creates 3 intervals: <math>(-\infty, -1)</math>, <math>(-1, 1)</math>, and <math>(1, \infty)</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
 
|-
 
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|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
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|Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
 
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|e) True.
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|<math>x = -2:</math> &nbsp;&nbsp;&nbsp; <math>\frac{-2 -1}{2(-2) + 2} = \frac{3}{2}</math>
 
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|f) False.  
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|<math>x = 0:</math> &nbsp;&nbsp;&nbsp; <math>\frac{-1}{2}</math>
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|-
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|<math>x = 2:</math> &nbsp;&nbsp;&nbsp; <math>\frac{2 - 1}{2(2) + 2} = \frac{1}{6}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 4: &nbsp;
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|-
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|The last check is whether or not the function intersects its horizontal asymptote. So check: <math>\frac{1}{2}=\frac{x - 1}{2x + 2}</math>.
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|- style = "text-align: center"
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|
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<math>\begin{array}{rcl}
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\frac{1}{2} &=& \frac{x - 1}{2x + 2}\\
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2x + 2 &=& 2(x - 1)\\
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2x + 2 &=& 2x - 2\\
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4 &=& 0
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\end{array}</math>
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|-
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|Since this is absurd, the function never intersects its horizontal asymptote. Now we graph while respecting the asymptotes
 
|}
 
|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Final Answer: &nbsp;
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|-
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|[[File:8A_Sample_Final_A,_Q_10.png]]
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|}
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[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]
 
[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

Revision as of 18:44, 2 June 2015

Question Consider the following rational function,


a. What is the domain of f?
b. What are the x and y-intercepts of f?
c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes?
d. Graph f(x). Make sure to include the information you found above.


Foundations:  
1) What are the asymptotes and zeros?
Answer:
1) The vertical asymptote corresponds to zeros of the denominator. So there is a vertical asymptote at x = -1. The zero is at (1, 0). The horizontal asymptote is the ratio of leading coefficients. So the horizontal asymptote is

Solution:

Step 1:  
We start by finding the asymptotes. The vertical asymptote corresponds to zeros of the denominator. So the vertical asymptote is at x = -1. They horizontal asymptote is determined by degree of the numerator and degree of the denominator. Since both of those values are 1, the horizontal asymptote is the ratio of leading coefficients. This means the horizontal asymptote is
Step 2:  
Now we observe that the zero is at (1, 0), and proceed by looking at the intervals created by removing x = -1 and x = 1. This creates 3 intervals: , , and
Step 3:  
Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
   
   
   
Step 4:  
The last check is whether or not the function intersects its horizontal asymptote. So check: .

Since this is absurd, the function never intersects its horizontal asymptote. Now we graph while respecting the asymptotes
Final Answer:  
8A Sample Final A, Q 10.png


Return to Sample Exam