008A Sample Final A, Question 10 - Revision history

MathAdmin at 06:56, 26 May 2015

2015-05-26T06:56:17Z

MathAdmin at 19:54, 23 May 2015

2015-05-23T19:54:18Z

MathAdmin at 19:53, 23 May 2015

2015-05-23T19:53:06Z

MathAdmin at 21:47, 28 April 2015

2015-04-28T21:47:24Z

MathAdmin: Created page with "'''Question: ''' Graph the function. Give equations of any asymptotes, and list any intercepts $y = \frac{x-1}{2x+2}$ {| class="mw-collapsible mw-collapsed" style..."

2015-04-28T21:10:54Z

Created page with "'''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..."

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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;"
!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 <math>y = \frac{1}{2}</math>
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! 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 <math>y = \frac{1}{2}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! 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: <math>(-\infty, -1)</math>, <math>(-1, 1)</math>, and <math>(1, \infty)</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
|-
|<math>x = -2:</math>     <math>\frac{-2 -1}{2(-2) + 2} = \frac{3}{2}</math>
|-
|<math>x = 0:</math>     <math>\frac{-1}{2}</math>
|-
|<math>x = 2:</math>     <math>\frac{2 - 1}{2(2) + 2} = \frac{1}{6}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
|-
|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>.
|- style = "text-align: center"
|
<math>\begin{array}{rcl}
\frac{1}{2} &=& \frac{x - 1}{2x + 2}\\
2x + 2 &=& 2(x - 1)\\
2x + 2 &=& 2x - 2\\
4 &=& 0
\end{array}</math>
|-
|Since this is absurd, the function never intersects its horizontal asymptote. Now we graph while respecting the asymptotes
|}

[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

← Older revision		Revision as of 19:54, 23 May 2015
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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>	+	'''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;"

← Older revision		Revision as of 19:53, 23 May 2015
Line 1:		Line 1:
−	'''Question: ''' Graph the function. Give equations of any asymptotes, and list any intercepts <math>y = \frac{x-1}{2x+2}</math>	+	'''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>

	{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"		{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations
+!Foundations: &nbsp;
 |-
 |1) What are the asymptotes and zeros?
@@ Line 14: / Line 14: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1:
+!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>
@@ Line 20: / Line 20: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2:
+!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>
@@ Line 26: / Line 26: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 3:
+!Step 3: &nbsp;
 |-
 |Now pick a number from each interval: -2, 0, 2 and find the value of the function for each number selected.
@@ Line 38: / Line 38: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 4:
+!Step 4: &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>.
@@ Line 54: / Line 54: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer:
+!Final Answer: &nbsp;
 |-
 |[[File:8A_Sample_Final_A,_Q_10.png]]