# Difference between revisions of "005 Sample Final A, Question 20"

Jump to navigation
Jump to search

(Created page with "'''Question ''' Consider the following rational function, <center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br> a. What is the domain of f?...") |
|||

(One intermediate revision by the same user not shown) | |||

Line 2: | Line 2: | ||

<center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br> | <center><math>f(x) = \frac{x^2+x-2}{x^2-1}</math></center> <br> | ||

− | + | :: a. What is the domain of f? <br> | |

− | + | :: b. What are the x and y-intercepts of f? <br> | |

− | + | :: c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes? <br> | |

− | + | :: d. Graph f(x). Make sure to include the information you found above. | |

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

− | ! | + | !Foundations: |

|- | |- | ||

− | | | + | |1) What points are not in the domain of f(x)? |

|- | |- | ||

− | | | + | |2) How do you find the intercepts? |

|- | |- | ||

− | | | + | |3) How do you find the asymptotes and zeros? |

|- | |- | ||

− | | | + | |4) How do you determine if f has any holes? |

|- | |- | ||

− | | | + | |Answer: |

|- | |- | ||

− | |f) | + | |1) The point that are not in the domain of f(x) are zeros of the denominator. |

+ | |- | ||

+ | |2) To find the x-intercepts set y = 0 and solve for x. For y-intercepts set x = 0 and simplify. | ||

+ | |- | ||

+ | |3) For zeros, find the zeros of the numerator. Vertical asymptotes correspond to zeros of the denominator. Horizontal asymptotes correspond to taking the limit as x goes to <math>\infty</math> | ||

+ | |- | ||

+ | |4) Holes occur when a single value of x is a zero of both the numerator and denominator. | ||

+ | |} | ||

+ | |||

+ | Solution: | ||

+ | |||

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

+ | !Step 1: | ||

+ | |- | ||

+ | |We start by finding the zeros of the denominator since this will give us information about vertical asymptotes and the domain. The zeros of the denominator are x = -1, 1. This tells us the domain is <math>(-\infty, -1) \cup (-1, 1) \cup (1, \infty)</math> and the potential vertical asymptotes are x = -1 and x = 1. | ||

+ | |} | ||

+ | |||

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

+ | !Step 2: | ||

+ | |- | ||

+ | |Now we should find the y-intercepts(zeros) to determine if f has any holes, which are zeros of numerator and denominator. Thus, y-intercepts correspond to zeros of the numerator x = -2, 1. Now we know we have a hole at x = 1, and a y-intercept at <math>(-2, 0).</math> | ||

|} | |} | ||

+ | |||

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

+ | !Step 3: | ||

+ | |- | ||

+ | |For the horizontal asymptote take the limit as x goes to <math>\infty.</math> This tells us that the horizontal asymptote is y = 1. | ||

+ | |} | ||

+ | |||

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

+ | !Final Answer: | ||

+ | |- | ||

+ | |Domain: <math>(-\infty, -1)\cup(-1, 1)\cup (1, \infty)</math> The x-intercept is <math>(-2, 0)</math> and y-intercept is <math>(0, 2)</math> | ||

+ | |- | ||

+ | |f does have a hole, with vertical asymptote at x = -1, and horizontal asymptote y = 1. | ||

+ | |- | ||

+ | |[[File:5_Sample_Final_20.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 19:18, 2 June 2015

**Question ** Consider the following rational function,

- a. What is the domain of f?
- b. What are the x and y-intercepts of f?
- c. What are the vertical and horizontal asymptotes of f, if any? Does f have any holes?
- d. Graph f(x). Make sure to include the information you found above.

- a. What is the domain of f?

Foundations: |
---|

1) What points are not in the domain of f(x)? |

2) How do you find the intercepts? |

3) How do you find the asymptotes and zeros? |

4) How do you determine if f has any holes? |

Answer: |

1) The point that are not in the domain of f(x) are zeros of the denominator. |

2) To find the x-intercepts set y = 0 and solve for x. For y-intercepts set x = 0 and simplify. |

3) For zeros, find the zeros of the numerator. Vertical asymptotes correspond to zeros of the denominator. Horizontal asymptotes correspond to taking the limit as x goes to |

4) Holes occur when a single value of x is a zero of both the numerator and denominator. |

Solution:

Step 1: |
---|

We start by finding the zeros of the denominator since this will give us information about vertical asymptotes and the domain. The zeros of the denominator are x = -1, 1. This tells us the domain is and the potential vertical asymptotes are x = -1 and x = 1. |

Step 2: |
---|

Now we should find the y-intercepts(zeros) to determine if f has any holes, which are zeros of numerator and denominator. Thus, y-intercepts correspond to zeros of the numerator x = -2, 1. Now we know we have a hole at x = 1, and a y-intercept at |

Step 3: |
---|

For the horizontal asymptote take the limit as x goes to This tells us that the horizontal asymptote is y = 1. |

Final Answer: |
---|

Domain: The x-intercept is and y-intercept is |

f does have a hole, with vertical asymptote at x = -1, and horizontal asymptote y = 1. |