Difference between revisions of "005 Sample Final A, Question 19"
Jump to navigation
Jump to search
(2 intermediate revisions by the same user not shown) | |||
Line 19: | Line 19: | ||
|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. | ||
|- | |- | ||
− | |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> | + | |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> |
|} | |} | ||
Line 31: | Line 31: | ||
|Amplitude: -1, period: <math>\frac{2\pi}{3}~</math>, phase shift: Left by <math>\frac{\pi}{2}~</math> and vertical shift up by 1. | |Amplitude: -1, period: <math>\frac{2\pi}{3}~</math>, phase shift: Left by <math>\frac{\pi}{2}~</math> and vertical shift up by 1. | ||
|} | |} | ||
− | |||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
! Step 2: | ! Step 2: | ||
|- | |- | ||
− | | | + | |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> |
− | | | + | |} |
− | + | ||
− | |- | + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" |
− | + | ! Step 3: | |
|- | |- | ||
− | | | + | |Now plot the five key points and sketch a graph that looks like the <math>\sin</math> graph through those points. |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
|} | |} | ||
Line 54: | Line 47: | ||
! Final Answer: | ! Final Answer: | ||
|- | |- | ||
− | | | + | |Amplitude: -1, period: <math>\frac{2\pi}{3}~</math>, phase shift: Left by <math>\frac{\pi}{2}~</math> and vertical shift up by 1. |
|- | |- | ||
− | |The | + | |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> |
|- | |- | ||
− | |[[File: | + | |[[File:5_Sample_Final_19.png]] |
|} | |} | ||
[[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.
- a. What is the amplitude?
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 |