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

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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.  
 
|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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|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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|2) The five key points are <math>(0, 0),~ (\frac{\pi}{2}, 1), ~ (\pi, 0), ~ (\frac{3\pi}{2}, 0),~ \text{and } (2\pi, 0).</math>
 
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! Step 1: &nbsp;
 
! Step 1: &nbsp;
 
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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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|We can read off the answers for (a) - (d):
 
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|Now we graph both functions.
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|Amplitude: -1, period: <math>\frac{2\pi}{3}~</math>, phase shift: Left by <math>\frac{\pi}{2}~</math> and vertical shift up by 1.
 
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! Step 2: &nbsp;
 
! Step 2: &nbsp;
 
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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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|The five key points for the graph are <math>(-\frac{\pi}{6}, 1), (0, 0), (\frac{\pi}{6}, 1), (\frac{\pi}{3}, 2), (\frac{\pi}{2}, 1)</math>
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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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|point satisfies the inequality or not. For both equations we will pick the origin.
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! Step 3: &nbsp;
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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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|<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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|<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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|We also shade the boundary of the circle since the inequality is <math>\le</math>
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|Now plot the five key points and sketch a graph that looks like the <math>\sin</math> graph through those points.
 
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! Final Answer: &nbsp;
 
! Final Answer: &nbsp;
 
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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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|Amplitude: -1, period: <math>\frac{2\pi}{3}~</math>, phase shift: Left by <math>\frac{\pi}{2}~</math> and vertical shift up by 1.
 
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|The region we are referring to is shaded both blue and red.
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|The five key points for the graph are <math>(-\frac{\pi}{6}, 1), (0, 0), (\frac{\pi}{6}, 1), (\frac{\pi}{3}, 2), (\frac{\pi}{2}, 1)</math>
 
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|[[File:8A_Final_5.png]]
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|[[File:5_Sample_Final_19.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 17:22, 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) The five key points are

Solution:

Step 1:  
We can read off the answers for (a) - (d):
Amplitude: -1, period: , phase shift: Left by and vertical shift up by 1.
Step 2:  
The five key points for the graph are
Step 3:  
Now plot the five key points and sketch a graph that looks like the graph through those points.
Final Answer:  
Amplitude: -1, period: , phase shift: Left by and vertical shift up by 1.
The five key points for the graph are
5 Sample Final 19.png


Return to Sample Exam

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