005 Sample Final A, Question 20 - Revision history

MathAdmin at 02:18, 3 June 2015

2015-06-03T02:18:02Z

MathAdmin at 01:44, 3 June 2015

2015-06-03T01:44:24Z

MathAdmin: Created page with "'''Question ''' Consider the following rational function, $f(x) = \frac{x^2+x-2}{x^2-1}$
a. What is the domain of f?..."

2015-06-01T01:38:25Z

Created page with "'''Question ''' Consider the following rational function, <center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> a. What is the domain of f?..."

New page

'''Question ''' Consider the following rational function,
<center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> 

     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.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answers
|-
|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>
|-
|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.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

[[005 Sample Final A|'''Return to Sample Exam''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|1) What are the asymptotes and zeros?
+|1) What points are not in the domain of f(x)?
 |-
 |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 <math>y = \frac{1}{2}</math>
+|1) The point that are not in the domain of f(x) are zeros of the denominator.
 |}
@@ Line 23: / Line 35: @@
 !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 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.
 |}
@@ Line 29: / Line 41: @@
 !Step 2: &nbsp;
 |-
-|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 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>
 |}
@@ Line 35: / Line 47: @@
 !Step 3: &nbsp;
 |-
-|Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
+|For the horizontal asymptote take the limit as x goes to <math>\infty.</math> This tells us that the horizontal asymptote is y = 1.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
+!Final Answer: &nbsp;
 |-
-|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>.
+|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>
 |-
-|Since this is absurd, the function never intersects its horizontal asymptote. Now we graph while respecting the asymptotes
+|f does have a hole, with vertical asymptote at x = -1, and horizontal asymptote y = 1.
 |-
-|[[File:8A_Sample_Final_A,_Q_10.png]]
+|[[File:5_Sample_Final_20.png]]
 |}
 [[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

@@ Line 2: / Line 2: @@
 <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>
+:: a. What is the domain of f? <br>
-&nbsp;&nbsp;&nbsp;&nbsp; b. What are the x and y-intercepts of f? <br>
+:: 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>
+:: 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.
+:: d. Graph f(x). Make sure to include the information you found above.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+!Foundations: &nbsp;
 |-
-|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>
+|1) What are the asymptotes and zeros?
 |-
-|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.
+|Answer:
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+|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>
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+|Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
 |-
-|e) True.
+|<math>x = -2:</math> &nbsp;&nbsp;&nbsp; <math>\frac{-2 -1}{2(-2) + 2} = \frac{3}{2}</math>
 |-
-|f) False.
+|<math>x = 0:</math> &nbsp;&nbsp;&nbsp; <math>\frac{-1}{2}</math>
 |}
 [[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

005 Sample Final A, Question 20 - Revision history

MathAdmin at 02:18, 3 June 2015

MathAdmin at 01:44, 3 June 2015

MathAdmin: Created page with "'''Question ''' Consider the following rational function, f(x) = \frac{x^2+x-2}{x^2-1} a. What is the domain of f?..."

MathAdmin: Created page with "'''Question ''' Consider the following rational function, $f(x) = \frac{x^2+x-2}{x^2-1}$
a. What is the domain of f?..."