quick vert. asymptote question (1 Viewer)

[ronaldinho]

the question is

(x^2 - 1) / (x^2 - 4)

find vert asymptote?

how do i do this one...

 

[redruM]

I'll have a stab.

If x = +/-2, the denominator will be equal to 0, hence y will be undefined for those values of x. Therefore, there are 2 vertical asymptotes at x = -2 and x = 2.

 

[pLuvia]

Allow the denominator to be zero, then solve for x, for horizontal just divide all the terms by teh highest x power and solve when x approaches infinity

 

[airie]

Just a note: make sure that the value of the whole expression DOES approach infinity (whether positive or negative) as x approaches the value(s) you found ie. the denominator becomes zero but the numerator doesn't. Sometimes a mere discontinuity occurs in the graph instead of an asymptote, for example at x=3 in the graph of (x²-9)/(x-3), which is simply the graph of y=x+3 with a "hole" at (3,6)