http://www.ck12.org Chapter 2. Polynomials and Rational Functions
- Draw the vertical asymptotes for the following function.
f(x) = 3 −(x+ 1 )(xx− 4 ) - Identify the holes and equations of the vertical asymptotes of the following rational function.
f(x) = (^53) ((xx+− 211 ))((x 2 ++^2 x)()(x 3 −−^3 x)()(xx+−^48 ))
Answers:
- Each criteria helps build the function. The vertical asymptotes imply that the denominator has two factors that
do not cancel with the numerator:
x·(x^1 − 3 )
The zeroes at 2 and 5 imply the numerator has two factors that do not cancel.
(x− 2 )(x− 5 )
x·(x− 3 )
The hole at (4, 2) implies that there is a factorx−4 that cancels on the numerator and the denominator.
(x− 2 )(x− 5 )(x− 4 )
x·(x− 3 )(x− 4 )
The tricky part is that the height of the function must be 2 after thex−4 factor has been canceled and the 4 is
substituted in. Currently it is−^12.
( 4 − 4 ·( 24 )(− 43 −) 5 )=− (^12)
In order to make the hole exist at a height of 2, you need to multiply the function by a scalar of -4.
f(x) =−^4 (xx·−(x^2 −)( 3 x)(−x^5 −)( 4 x)−^4 )
- The vertical asymptotes occur atx=− 1 ,x=4.
- The vertical asymptotes occur atx=−^12 ,x=8. Holes occur whenxis -2 and 3. To get the height of the holes
at these points, remember to cancel what can be canceled and then substitute the values. A very common mistake is
to forget to cancelx 3 −−^3 x=−1.
g(x) =− 53 ((xx+− 11 )(x+^4 )
2)(x− 8 )g(− 2 ) = 256
g( 3 ) =^1225