# Thread: a quick question about a rational polynomial and it's graph.

1. ## a quick question about a rational polynomial and it's graph.

so im in taking a precalc college course. for homework i was given the rational polynomial x/x(x+4) and told to graph it. when it comes to graphing polynomials (both rational or otherwise) i check my answers in an online graphing caculator. it was clear to me that the graphing caculator had simplified the equation to 1/x+4, and hence we (the caculator and i) came up with different graphs. i simplified the rational polynomial, and it all made sense (the caculator and i were in agreement). but im still somewhat confused. why should that be the case? why should a rational polynomial and its simplified form give different graphs?

to be more specific....

i see why there would be no x intercept just by looking at the original unsimplified polynomial. set the rational polynomial equal to zero, we get an x intercept of (0,0), however setting x to 0 would make the rational polynomial undefined, so there is no x intecept (this is exactly what the simplified form suggests). they also give the same horizontal asymtote, namely y=0. however the unsimplified form also gives no y intercept, and the simplified form does, namely (0,0.25). furthermore, they both give different vertical asymtotes: simplified gives x=-4 and unsimplified gives x=-4 and x=0. there has to be some way to logically think about the unsimplified such that they both give the same graph. or maybe there is some mathematical proof? i dunno, this isn't exactly urgent, but some insight would be appreciated.

i should also note i brought this up to my proffessor when he was going over the question. he was as confused as i was, and proceeded to solve the problem in a way the caculator would disagree with.

2. ## Re: a quick question about a rational polynomial and it's graph.

the graph of $\displaystyle y = \frac{x}{x(x+4)}$ is identical to the graph $\displaystyle y = \frac{1}{x+4}$ except at $\displaystyle x = 0$

the first function has a point discontinuity at x = 0 , y = 1/4

the second function has the "hole" at x = 0 , y = 1/4 filled in, and is continuous at x = 0.

the graph attached is $\displaystyle y = \frac{x}{x(x+4)}$ in a decimal window (TI-84) with the axes turned off ... see the "hole" ?

3. ## Re: a quick question about a rational polynomial and it's graph.

hi skeeter, sry it took so long to respond.

i cant say i completley understand. i have never heard the term point discontinuity used before, or the "hole". is some of this beyond the realm of precalc? anyways, trying to explain this to me may be a lost cause, feel free to tell me to buzz off.

4. ## Re: a quick question about a rational polynomial and it's graph.

Originally Posted by inaudibletree
hi skeeter, sry it took so long to respond.

i cant say i completley understand. i have never heard the term point discontinuity used before, or the "hole". is some of this beyond the realm of precalc? anyways, trying to explain this to me may be a lost cause, feel free to tell me to buzz off.
this concept is usually taught in an algebra 2 course (before precalculus) ... and your "prof" was confused?

5. ## Re: a quick question about a rational polynomial and it's graph.

thanks for posting the vid skeeter. i redid the graphs for both functions and compared them, defintley makes sense now. i cant believe my professor solved this wrong....he is apparently qualified to teach caculus....