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

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!Foundations:  
 
!Foundations:  
 
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|1) What are the asymptotes and zeros?
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|1) What points are not in the domain of f(x)?
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|2) How do you find the intercepts?
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|3) How do you find the asymptotes and zeros?
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|4) How do you determine if f has any holes?
 
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|Answer:
 
|Answer:
 
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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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|1) The point that are not in the domain of f(x) are zeros of the denominator.
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|-
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|2) To find the x-intercepts set y = 0 and solve for x. For y-intercepts set x = 0 and simplify.
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|3) For zeros, find the zeros of the numerator. Vertical asymptotes correspond to zeros of the denominator. Horizontal asymptotes correspond to taking the limit as x goes to <math>\infty</math>
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|4) Holes occur when a single value of x is a zero of both the numerator and denominator.
 
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!Step 1: &nbsp;
 
!Step 1: &nbsp;
 
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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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|We start by finding the zeros of the denominator since this will give us information about vertical asymptotes and the domain. The zeros of the denominator are x = -1, 1. This tells us the domain is <math>(-\infty, -1) \cup (-1, 1) \cup (1, \infty)</math> and the potential vertical asymptotes are x = -1 and x = 1.
 
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!Step 2: &nbsp;
 
!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>
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|Now we should find the y-intercepts(zeros) to determine if f has any holes, which are zeros of numerator and denominator. Thus, y-intercepts correspond to zeros of the numerator x = -2, 1. Now we know we have a hole at x = 1, and a y-intercept at <math>(-2, 0).</math>
 
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!Step 3: &nbsp;
 
!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.
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|For the horizontal asymptote take the limit as x goes to <math>\infty.</math> This tells us that the horizontal asymptote is y = 1.
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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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|<math>x = 0:</math> &nbsp;&nbsp;&nbsp; <math>\frac{-1}{2}</math>
 
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 4: &nbsp;
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!Final Answer: &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>.
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|Domain: <math>(-\infty, -1)\cup(-1, 1)\cup (1, \infty)</math> The x-intercept is <math>(-2, 0)</math> and y-intercept is <math>(0, 2)</math>
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|- style = "text-align: center"
 
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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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|Since this is absurd, the function never intersects its horizontal asymptote. Now we graph while respecting the asymptotes
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|f does have a hole, with vertical asymptote at x = -1, and horizontal asymptote y = 1.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
 
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|[[File:8A_Sample_Final_A,_Q_10.png]]
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|[[File:5_Sample_Final_20.png]]
 
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[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]
 
[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 18:18, 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 points are not in the domain of f(x)?
2) How do you find the intercepts?
3) How do you find the asymptotes and zeros?
4) How do you determine if f has any holes?
Answer:
1) The point that are not in the domain of f(x) are zeros of the denominator.
2) To find the x-intercepts set y = 0 and solve for x. For y-intercepts set x = 0 and simplify.
3) For zeros, find the zeros of the numerator. Vertical asymptotes correspond to zeros of the denominator. Horizontal asymptotes correspond to taking the limit as x goes to
4) Holes occur when a single value of x is a zero of both the numerator and denominator.

Solution:

Step 1:  
We start by finding the zeros of the denominator since this will give us information about vertical asymptotes and the domain. The zeros of the denominator are x = -1, 1. This tells us the domain is and the potential vertical asymptotes are x = -1 and x = 1.
Step 2:  
Now we should find the y-intercepts(zeros) to determine if f has any holes, which are zeros of numerator and denominator. Thus, y-intercepts correspond to zeros of the numerator x = -2, 1. Now we know we have a hole at x = 1, and a y-intercept at
Step 3:  
For the horizontal asymptote take the limit as x goes to This tells us that the horizontal asymptote is y = 1.
Final Answer:  
Domain: The x-intercept is and y-intercept is
f does have a hole, with vertical asymptote at x = -1, and horizontal asymptote y = 1.
5 Sample Final 20.png


Return to Sample Exam

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