• Nov 15th 2009, 04:04 PM
dcd5105
Having a bit of trouble grasping how to define vertical asymptotes and holes...

\$\displaystyle f(x)=x^2-25/x^2+5x\$

\$\displaystyle f(x)=(x-5)(x+5)/x(x+5)\$
\$\displaystyle f(x)=x-5/x\$

I get that since the leading coefficients are both 1 and the degrees are the same, the horizontal asymptote is \$\displaystyle y=1\$. Can someone please help me out and give me a breakdown of defining the vertical asymptote and any holes there might be? Any help is greatly appreciated!
• Nov 15th 2009, 04:08 PM
Jhevon
Quote:

Originally Posted by dcd5105
Having a bit of trouble grasping how to define vertical asymptotes and holes...

\$\displaystyle f(x)=x^2-25/x^2+5x\$

\$\displaystyle f(x)=(x-5)(x+5)/x(x+5)\$
\$\displaystyle f(x)=x-5/x\$

I get that since the leading coefficients are both 1 and the degrees are the same, the horizontal asymptote is \$\displaystyle y=1\$. Can someone please help me out and give me a breakdown of defining the vertical asymptote and any holes there might be? Any help is greatly appreciated!

Vertical asymptotes occur when the function is undefined. In the case of rational functions, this is where the denominator is zero.

In the special case that the numerator is zero at the same point, due to the same factor, you get a hole. for example, in your first function, you have a vertical asymptote at x = 0 (the denominator is zero when x = 0, but not the numerator), but a hole at x = -5 (since both the denominator and the numerator are zero there, you can factor the top and bottom in such a way as to "cancel" the bad factor).