Difference between revisions of "008A Sample Final A, Question 10"

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'''Question: '''  Graph the function. Give equations of any asymptotes, and list any intercepts <math style = "vertical-align:baseline">y = \frac{x-1}{2x+2}</math>
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'''Question: '''  Graph the function. Give equations of any asymptotes, and list any intercepts <math>y = \frac{x-1}{2x+2}</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
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!Foundations: &nbsp;
 
|-
 
|-
 
|1) What are the asymptotes and zeros?
 
|1) What are the asymptotes and zeros?
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
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!Step 1: &nbsp;
 
|-
 
|-
 
|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>
 
|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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
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!Step 2: &nbsp;
 
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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>
 
|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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
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!Step 3: &nbsp;
 
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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.
 
|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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
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!Step 4: &nbsp;
 
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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>.
 
|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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
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!Final Answer: &nbsp;
 
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|[[File:8A_Sample_Final_A,_Q_10.png]]
 
|[[File:8A_Sample_Final_A,_Q_10.png]]

Latest revision as of 23:56, 25 May 2015

Question: Graph the function. Give equations of any asymptotes, and list any intercepts

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

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