**T O P I C R E V I E W** |

**effort** |
Posted - 12/05/2012 : 12:19:31 Here is the problem; f(x) = (x)/(x-x-2)
I know that the vertical asymptotes are x = -1 and x= 2 and the horizontal asymptote is y = 0 My problem is I can't get my kid to understand that y = 0 is an asymptote. She doesn't understand why that middle piece of the graph does crosses y = 0 if y = 0 is an asymptote. Can someone help me explain why ? |

**2 L A T E S T R E P L I E S (Newest First)** |

**Subhotosh Khan** |
Posted - 12/06/2012 : 15:31:19 quote:
*Originally posted by effort*
Here is the problem; f(x) = (x)/(x-x-2)
I know that the vertical asymptotes are x = -1 and x= 2 and the horizontal asymptote is y = 0 My problem is I can't get my kid to understand that y = 0 is an asymptote. She doesn't understand why that middle piece of the graph does crosses y = 0 if y = 0 is an asymptote. Can someone help me explain why ?
A vertical asymptote of a function will __not__ have the graph crossing the assymptote (due to the requirement of vertical-line tests for functions - or - a function can have only one value at one 'x').
However, there is no such restriction for horizontal line in a function (A function can have same value at multiple x's) |

**Ultraglide** |
Posted - 12/05/2012 : 17:43:06 For vertical asymptotes, the function does not cross the asymptotic line. For horizontal ones, it may. What I usually do is use a couple of values of x in each direction, i.e. 10 and 100 for the right and -10 and -100 for the left. For the first two you get 10/88 and 100/9898 which give approximately 0.11 and 0.01 which gets close to zero from above. For the second two you get -10/108 and -100/10098 which give -0.09 and -0.01 (also approximations). This approaches zero from below because the values are negative. You can try values even further out from zero and will get even closer to zero from above and below. Therefore the horizontal asymptote is indeed zero. If you want to graph the function, use a free graphing program such as Graphmatica if you don't have a graphing calculator. If you look at the first derivative, you will find that the function is continually decreasing as it approaches positive infinity or inreasing as x approaches negative infinity. The fact that the function crosses the asymptote makes no difference. |