# Difference between revisions of "008A Sample Final A, Question 3"

Jump to navigation
Jump to search

Line 9: | Line 9: | ||

! Foundations | ! Foundations | ||

|- | |- | ||

− | |1) What type of | + | |1) What type of graph is this? (line, parabola, circle, etc.) |

|- | |- | ||

|2) What is the process for transforming the function into the standard graphing form? | |2) What is the process for transforming the function into the standard graphing form? | ||

Line 38: | Line 38: | ||

! Step 1: | ! Step 1: | ||

|- | |- | ||

− | |There are two ways to | + | |There are two ways to determine the standard graphing form. |

|- | |- | ||

|Regardless of the method the first step is the same: subtract 2 from both sides to yield <math>x - 2 = -3y^2 - 6y</math> | |Regardless of the method the first step is the same: subtract 2 from both sides to yield <math>x - 2 = -3y^2 - 6y</math> | ||

Line 55: | Line 55: | ||

|- | |- | ||

|Now we complete the square inside the parenthesis and add -3 to the left hand side resulting in <math>x - 5 = -3(y^2 + 2y +1) = -3(y + 1)^2</math> | |Now we complete the square inside the parenthesis and add -3 to the left hand side resulting in <math>x - 5 = -3(y^2 + 2y +1) = -3(y + 1)^2</math> | ||

+ | |- | ||

+ | |We mention here that some instructors/professors are particular about which of these two methods you use. | ||

|} | |} | ||

Line 60: | Line 62: | ||

! Step 2: | ! Step 2: | ||

|- | |- | ||

− | |Since the parabola is in standard graphing form we can read off the vertex, which is ( | + | |Since the parabola is in standard graphing form we can read off the vertex, which is (5, -1). |

|- | |- | ||

|We get the X-intercept by replacing y with 0 and solving for x. So <math>x - 5 = -3(1)^2</math>, and the X-intercept is (2, 0). | |We get the X-intercept by replacing y with 0 and solving for x. So <math>x - 5 = -3(1)^2</math>, and the X-intercept is (2, 0). | ||

Line 76: | Line 78: | ||

! Step 4: | ! Step 4: | ||

|- | |- | ||

− | |Since <math>a < 0</math>, the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at <math>( | + | |Since <math>a < 0</math>, the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at <math>(5 - \frac{1}{12}, -1) = (\frac{59}{12}, -1)</math>. |

|- | |- | ||

− | |We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is <math>x = | + | |We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is <math>x = 5 + \frac{1}{12} = \frac{61}{2}</math> |

|} | |} | ||

Line 84: | Line 86: | ||

! Final Answer: | ! Final Answer: | ||

|- | |- | ||

− | |Vertex: ( | + | |Vertex: (5, -1), standard graphing form: <math>x - 5 = -3(y + 1)^2</math>, X-intercept: (2, 0), focus: <math>(\frac{59}{12}, -1)</math>, directrix: <math>x = \frac{61}{2}</math> |

[[File:8A_Sample_Final_A,_Q_3.png]] | [[File:8A_Sample_Final_A,_Q_3.png]] | ||

|} | |} | ||

[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']] | [[008A Sample Final A|<u>'''Return to Sample Exam</u>''']] |

## Revision as of 23:08, 22 May 2015

**Question:** a) Find the vertex, standard graphing form, and X-intercept for

b) Sketch the graph. Provide the focus and directrix.

Note: In this problem, what is referred to as standard graphing form is the vertex form, in case you search on the internet.

Foundations |
---|

1) What type of graph is this? (line, parabola, circle, etc.) |

2) What is the process for transforming the function into the standard graphing form? |

3) After we have the standard graphing form how do you find the X-intercept, and vertex? |

4) Moving on to part b) How do we find a point on the graph? |

5) From the standard graphing form how do we obtain relevant information about the focus and directrix? |

Answers: |

1) The function is a parabola. Some of the hints: We are asked to find the vertex, and directrix. Also only one variable, of x and y, is squared. |

2) First we complete the square. Then we divide by the coefficient of x. |

3) To find the X-intercept, replace y with 0 and solve for x. Since the parabola is in standard graphing form, the vertex of is (h, k). |

4) To find a point, we can either use the symmetry of a parabola or plug in another value for x. |

5) From the equation , we use the equation to find p. P is both the distance from the vertex to the focus and the distance from the vertex to the directrix. |

Solution:

Step 1: |
---|

There are two ways to determine the standard graphing form. |

Regardless of the method the first step is the same: subtract 2 from both sides to yield |

Method 1: |

Divide both sides by -3 to make the coefficient of , 1. This means |

Complete the square to get |

Multiply both sides by -3 so , and simplify the left side to yield |

Method 2: |

Instead of dividing by -3 we factor it out of the right hand side to get . |

Now we complete the square inside the parenthesis and add -3 to the left hand side resulting in |

We mention here that some instructors/professors are particular about which of these two methods you use. |

Step 2: |
---|

Since the parabola is in standard graphing form we can read off the vertex, which is (5, -1). |

We get the X-intercept by replacing y with 0 and solving for x. So , and the X-intercept is (2, 0). |

Step 3: |
---|

Now we need the value of p using the relation , where a = -3. |

So , and . |

Step 4: |
---|

Since , the parabola opens left. Since the focus is inside the parabola, and p tells us the focus is from the vertex, the focus is at . |

We also know that the directrix is a vertical line on the outside of the parabola with the distance from the directrix to the vertex being p. Thus the directrix is |

Final Answer: |
---|

Vertex: (5, -1), standard graphing form: , X-intercept: (2, 0), focus: , directrix: |