Difference between revisions of "005 Sample Final A, Question 19"

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(Created page with "'''Question ''' Consider the following function, <center><math>f(x) = -\sin\left(3x+\frac{\pi}{2}\right)+1</math></center><br>      a. What is the amplitu...")
 
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'''Question ''' Consider the following function, <center><math>f(x) = -\sin\left(3x+\frac{\pi}{2}\right)+1</math></center><br>
 
'''Question ''' Consider the following function, <center><math>f(x) = -\sin\left(3x+\frac{\pi}{2}\right)+1</math></center><br>
  
&nbsp;&nbsp;&nbsp;&nbsp; a. What is the amplitude?<br>
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:: a. What is the amplitude?<br>
&nbsp;&nbsp;&nbsp;&nbsp; b. What is the period? <br>
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:: b. What is the period? <br>
&nbsp;&nbsp;&nbsp;&nbsp; c. What is the phase shift? <br>
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:: c. What is the phase shift? <br>
&nbsp;&nbsp;&nbsp;&nbsp; d. What is the vertical shift? <br>
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:: d. What is the vertical shift? <br>
&nbsp;&nbsp;&nbsp;&nbsp; e. Graph one cycle of f(x). Make sure to label five key points.
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:: e. Graph one cycle of f(x). Make sure to label five key points.
  
  
 
{| 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) For parts (a) - (d), How do we read the relevant information off of <math>A\sin(Bx + C) + D?</math>
 
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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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|2) What are the five key points when looking at <math>\sin(x)?</math>
 
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|c) False. <math>y = x^2</math> does not have an inverse.
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|Answer:
 
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|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
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|1) The amplitude is A, the period is <math>\frac{2\pi}{B}</math>, the horizontal shift is left by C units if C is positive and right by C units if C is negative, the vertical shift is up by D if D is positive and down by D units if D is negative.  
 
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|e) True.
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|2) Since the Y-value must be less than <math>\vert x\vert + 1</math>, shade below the V. For the circle shde the inside.
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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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|First we replace the inequalities with equality. So <math>y = \vert x\vert + 1</math>, and <math>x^2 + y^2 = 9</math>.
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|-
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|Now we graph both functions.
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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 that we have graphed both functions we need to know which region to shade with respect to each graph.
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|-
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|To do this we pick a point an equation and a point not on the graph of that equation. We then check if the
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|-
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|point satisfies the inequality or not. For both equations we will pick the origin.
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|-
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|<math>y < \vert x\vert + 1:</math> Plugging in the origin we get, <math>0 < \vert 0\vert + 1 = 1</math>. Since the inequality is satisfied shade the side of
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|-
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|<math>y < \vert x\vert + 1</math> that includes the origin. We make the graph of <math>y < \vert x\vert + 1</math>, since the inequality is strict.
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|-
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|<math>x^2 + y^2 \le 9:</math> <math>(0)^2 +(0)^2 = 0 \le 9</math>. Once again the inequality is satisfied. So we shade the inside of the circle.
 
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|f) False.
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|We also shade the boundary of the circle since the inequality is <math>\le</math>
 
|}
 
|}
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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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|The final solution is the portion of the graph that below <math>y = \vert x\vert + 1</math> and inside <math> x^2 + y^2 = 9</math>
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|-
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|The region we are referring to is shaded both blue and red.
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|-
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|[[File:8A_Final_5.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 12:05, 2 June 2015

Question Consider the following function,


a. What is the amplitude?
b. What is the period?
c. What is the phase shift?
d. What is the vertical shift?
e. Graph one cycle of f(x). Make sure to label five key points.


Foundations:  
1) For parts (a) - (d), How do we read the relevant information off of
2) What are the five key points when looking at
Answer:
1) The amplitude is A, the period is , the horizontal shift is left by C units if C is positive and right by C units if C is negative, the vertical shift is up by D if D is positive and down by D units if D is negative.
2) Since the Y-value must be less than , shade below the V. For the circle shde the inside.

Solution:

Step 1:  
First we replace the inequalities with equality. So , and .
Now we graph both functions.
Step 2:  
Now that we have graphed both functions we need to know which region to shade with respect to each graph.
To do this we pick a point an equation and a point not on the graph of that equation. We then check if the
point satisfies the inequality or not. For both equations we will pick the origin.
Plugging in the origin we get, . Since the inequality is satisfied shade the side of
that includes the origin. We make the graph of , since the inequality is strict.
. Once again the inequality is satisfied. So we shade the inside of the circle.
We also shade the boundary of the circle since the inequality is
Final Answer:  
The final solution is the portion of the graph that below and inside
The region we are referring to is shaded both blue and red.
8A Final 5.png


Return to Sample Exam