# Difference between revisions of "005 Sample Final A, Question 16"

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(Created page with "'''Question ''' Graph the following, <center><math> -x^2+4y^2-2x-16y+11=0</math></center> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Final Answers |-...") |
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||

− | ! | + | !Foundations: |

|- | |- | ||

− | | | + | |1) What type of function is this? |

|- | |- | ||

− | | | + | |2) What can you say about the orientation of the graph? |

|- | |- | ||

− | | | + | |Answer: |

|- | |- | ||

− | | | + | |1) Since both x and y are squared it must be a hyperbola or an ellipse. We can conclude that the graph is an ellipse since both <math>x^2</math> and <math>y^2</math> have the same sign, positive. |

|- | |- | ||

− | | | + | |2) Since the coefficient of the <math>x^2</math> term is smaller, when we divide both sides by 36 the X-axis will be the major axis. |

+ | |} | ||

+ | |||

+ | Solution: | ||

+ | |||

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

+ | ! Step 1: | ||

|- | |- | ||

− | | | + | |We start by dividing both sides by 36. This yields <math>\frac{4x^2}{36} + \frac{9(y + 1)^2}{36} = \frac{x^2}{9} + \frac{(y+1)^2}{4} = 1</math>. |

+ | |} | ||

+ | |||

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

+ | ! Step 2: | ||

+ | |- | ||

+ | |Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1). | ||

+ | |- | ||

+ | |From the center mark the two points that are 3 units left, and three units right of the center. | ||

+ | |- | ||

+ | |Then mark the two points that are 2 units up, and two units down from the center. | ||

+ | |} | ||

+ | |||

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

+ | ! Final Answer: | ||

+ | |- | ||

+ | |The four vertices are: <math>(-3, -1), (3, -1), (0, 1) \text{ and } (0, -3)</math> | ||

+ | |- | ||

+ | |[[File:8A_Sample_Final,_Q_6.png]] | ||

|} | |} | ||

[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']] | [[005 Sample Final A|'''<u>Return to Sample Exam</u>''']] |

## Revision as of 10:59, 2 June 2015

**Question ** Graph the following,

Foundations: |
---|

1) What type of function is this? |

2) What can you say about the orientation of the graph? |

Answer: |

1) Since both x and y are squared it must be a hyperbola or an ellipse. We can conclude that the graph is an ellipse since both and have the same sign, positive. |

2) Since the coefficient of the term is smaller, when we divide both sides by 36 the X-axis will be the major axis. |

Solution:

Step 1: |
---|

We start by dividing both sides by 36. This yields . |

Step 2: |
---|

Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1). |

From the center mark the two points that are 3 units left, and three units right of the center. |

Then mark the two points that are 2 units up, and two units down from the center. |

Final Answer: |
---|

The four vertices are: |