# graph asymptote question

• Sep 24th 2006, 09:40 PM
scorpion007
graph asymptote question
hi, im having a problem with this question. Im not sure how even go about getting one of these multiple choices.

The graph of $f(x) = \frac{1}{x^2-mx -n}$, where m and n are real constants, has no vertical asymptotes if

A: $m^2 < 4n$
B: $m^2 > 4n$
C: $m^2 = -4n$
D: $m^2 < -4n$
E: $m^2 > -4n$

help appreciated.
• Sep 24th 2006, 10:46 PM
CaptainBlack
Quote:

Originally Posted by scorpion007
hi, im having a problem with this question. Im not sure how even go about getting one of these multiple choices.

The graph of $f(x) = \frac{1}{x^2-mx -n}$, where m and n are real constants, has no vertical asymptotes if

A: $m^2 < 4n$
B: $m^2 > 4n$
C: $m^2 = -4n$
D: $m^2 < -4n$
E: $m^2 > -4n$

help appreciated.

<obligatory moan about LaTeX not being working yet>

For f to have a vertical asymtote requires that x^2-mx-n have at least
one real root. This is equivalent to the discriminanat being >=0.

formula, so in this case the disciminant is: m^2+4n, and so f has vetical
asymtots if (and only if):

m^2+4n >=0,

or:

m^2 >= -4n.

RonL
• Sep 25th 2006, 02:49 AM
scorpion007
hah, sorry about the latex, i wont do it next time.

Also -- i found the answer myself to the problem shortly after posting, lol.
BTW, the question asks for the opposite of what you found. i.e. when the graph has *no* vert asymptotes. But all you have to do is change the >= to <.

Thanks anyway