# 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) | + | |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: | ! Step 1: | ||

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− | | | + | |We can read off the answers for (a) - (d): |

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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. |

|} | |} | ||

+ | |||

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

## Revision as of 12:43, 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: |
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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: |
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We can read off the answers for (a) - (d): |

Amplitude: -1, period: , phase shift: Left by and vertical shift up by 1. |

Step 2: |
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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: |
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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. |