http://www.ck12.org Chapter 2. Polynomials and Rational Functions
The removable discontinuity occurs at (1, 5).
Example C
Graph the following rational function and identify any removable discontinuities.
f(x) =x^6 −^6 x^5 +^5 x^4 x+ (^327) − 7 xx^3 −− 648 x^2 −^9 x+^54
Solution: This is probably one of the most challenging rational expressions with only holes that people ever try to
graph by hand. There are multiple ways to start, but a good habit to get into is to factor everything you possibly
can initially. The denominator seems less complicated with possible factors(x± 1 ),(x± 2 ),(x± 3 ),(x± 6 ). Using
polynomial division, you will find the denominator becomes:
f(x) =x^6 −^6 x^5 +(x^5 +x^41 +)(^27 x+x 23 −)(x^48 −x 32 )−^9 x+^54
The factors of the denominator are strong hints as to the factors of the numerator so use polynomial division and try
each. When you fully factor the numerator you will have:
f(x) =(x^3 −^6 x^2 +(x^12 + 1 x−)(^9 x)(+ 2 x+)(^1 x)(− 3 x+)^2 )(x−^3 )
Note the factors that cancel(x=− 1 ,− 2 , 3 )and then work with the cubic function that remains.
f(x) =x^3 − 6 x^2 + 12 x− 9
At this point it is probably reasonable to make a table and plot points to get a sense of where this cubic function
lives. You also could notice that the coefficients are almost of the pattern 1 3 3 1 which is the binomial expansion. By
separating the -9 into -8 -1 you can factor the first four terms.
f(x) =x^3 − 6 x^2 + 12 x− 8 − 1 = (x− 2 )^3 − 1
This is a cubic function that has been shifted right by two units and down one unit.