# 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> | ||

− | + | :: a. What is the amplitude?<br> | |

− | + | :: b. What is the period? <br> | |

− | + | :: c. What is the phase shift? <br> | |

− | + | :: d. What is the vertical shift? <br> | |

− | + | :: 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;" | ||

− | ! | + | ! Foundations: |

|- | |- | ||

− | |a) | + | |1) For parts (a) - (d), How do we read the relevant information off of <math>A\sin(Bx + C) + D?</math> |

|- | |- | ||

− | | | + | |2) What are the five key points when looking at <math>\sin(x)?</math> |

|- | |- | ||

− | | | + | |Answer: |

|- | |- | ||

− | | | + | |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) Since the Y-value must be less than <math>\vert x\vert + 1</math>, shade below the V. For the circle shde the inside. |

+ | |} | ||

+ | |||

+ | Solution: | ||

+ | |||

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

+ | ! Step 1: | ||

+ | |- | ||

+ | |First we replace the inequalities with equality. So <math>y = \vert x\vert + 1</math>, and <math>x^2 + y^2 = 9</math>. | ||

+ | |- | ||

+ | |Now we graph both functions. | ||

+ | |} | ||

+ | |||

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

+ | ! 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. | ||

+ | |- | ||

+ | |<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 | ||

+ | |- | ||

+ | |<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. | ||

+ | |- | ||

+ | |<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. | ||

|- | |- | ||

− | | | + | |We also shade the boundary of the circle since the inequality is <math>\le</math> |

|} | |} | ||

+ | |||

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

+ | ! Final Answer: | ||

+ | |- | ||

+ | |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> | ||

+ | |- | ||

+ | |The region we are referring to is shaded both blue and red. | ||

+ | |- | ||

+ | |[[File:8A_Final_5.png]] | ||

+ | |} | ||

+ | |||

[[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.

- 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) 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. |