How do you graph #f(x)=8/(x(x+2))# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Dec 19, 2017

holes: a value that causes both the numerator and denominator to equal zero. there are no holes in this rational function.

vertical asymptotes: it's a line #-># set the denominator of the rational function equal to #0#:

#x(x+2) = 0#
vertical asymptotes: #x = 0, x = -2#

horizontal asymptotes:
the following are the rules for solving horizontal asymptotes:
let m be the degree of the numerator
let n be the degree of the denominator

if m > n, then there is no horizontal asymptote

if m = n, then the horizontal asymptote is dividing the coefficients of the numerator and denominator

if m < n, then the horizontal asymptote is #y = 0#.

As we can see in our rational function, the denominator has a larger degree of #x#. So the horizontal asymptote is #y = 0#.

x-ints: x-intercepts are the top of the rational function. Since the numerator just says #8#, that means that there are no x-ints.

y-ints: y-intercepts are when you plug in #0# to the function:
#8/(0(0+2))#
#8/0 -> # undefined, so there are no y-ints.

Hope this helps!