2.13. Graphs of Rational Functions by Hand http://www.ck12.org
Solution: After attempting to factor the numerator you may realize that bothx=1 andx=−2 are vertical
asymptotes rather than holes. The horizontal asymptote isy=^12. There are no holes. They-intercept is:
f( 0 ) =− 81 · 2 = 161
The numerator is not factorable, but there is a zero between 0 and 1. You know this because there are no holes or
asymptotes between 0 and 1 and the function switches from negative to positive in this region.
4 ( 0 )^3 − 2 ( 0 )^2 + 3 ( 0 )− 1 =− 1
4 ( 1 )^3 − 2 ( 1 )^2 + 3 ( 1 )− 1 = 4
Putting all of this together in a sketch:
Example B
Completely plot the following rational function.
f(x) =(x−^3 )(^2) (x− 2 ) (^3) (x− 1 )(x+ 2 )
300 (x+ 1 )^2 (x− 2 )x
Solution: Since this function is already factored, much of the work is already done. There is a hole at (2, 0). There
are two vertical asymptotes atx=− 1 ,0. There are no horizontal or oblique asymptotes because the degree of the
numerator is much bigger than the degree of the denominator. Asxgets large this function grows without bound. As
xget very small, this function decreases without bound. The function has noyintercept because that is where a
vertical asymptote is. Besides the hole at (2, 0), there are zeroes at (-2, 0), (1, 0) and (3, 0). This is what the graph
ends up looking like.