A rational function $y=\frac{P\left(x\right)}{Q\left(x\right)}$y=P(x)Q(x) where the degree of $P\left(x\right)$P(x) is greater than or equal to $Q\left(x\right)$Q(x), can be re-expressed by division as:
$y=\frac{P\left(x\right)}{Q\left(x\right)}=D\left(x\right)+\frac{R\left(x\right)}{Q\left(x\right)}$y=P(x)Q(x)=D(x)+R(x)Q(x)
where $D\left(x\right)$D(x) is known as the divisor polynomial and $R\left(x\right)$R(x) is known as the remainder polynomial.
By re-expressing in this way, certain features of the graph become evident - particularly the behaviour of the curve at extreme points.
Exploration
Example 1
For example, the rational function given by $y=\frac{x^2+2}{x}$y=x2+2x can be re-expressed as $y=x+\frac{2}{x}$y=x+2x. hence we can see that the rational function is the sum of the linear function $y=x$y=x and the hyperbolic function $y=\frac{2}{x}$y=2x.
The new form also makes it clear that, as $x\rightarrow\infty$x→∞, the quantity $\frac{2}{x}\rightarrow0$2x→0 and thus the function values become asymptotically closer from above to the line $y=x$y=x. In addition, as $x\rightarrow-\infty$x→−∞, the curve approaches the line $y=x$y=x asymptotically from below.
We also see that as $x\rightarrow0$x→0 from above (for example $x=1$x=1, $x=0.5$x=0.5, $x=0.1$x=0.1, $x=0.01$x=0.01 etc), the quantity $\frac{2}{x}\rightarrow\infty$2x→∞. The same type of behaviour occurs when $x\rightarrow0$x→0 from below (for example $x=-1$x=−1, $x=-0.5$x=−0.5, $x=-0.1$x=−0.1, $x=-0.01$x=−0.01 etc) the quantity $\frac{2}{x}\rightarrow-\infty$2x→−∞. These two results mean that the curve approaches the vertical asymptote $x=0$x=0 in opposite directions.
The graph below is now much easier to interpret in terms of this re-expressed form. The extreme behaviour and the behaviour around $x=0$x=0 makes sense.
Example 2
As another example, the rational function given by $y=\frac{x^3-1}{x^3-2}$y=x3−1x3−2 can be re-expressed as $y=1+\frac{1}{x^3-2}$y=1+1x3−2 by noting that $\frac{x^3-1}{x^3-2}=\frac{\left(x^3-2\right)+1}{x^3-2}=1+\frac{1}{x^3-2}$x3−1x3−2=(x3−2)+1x3−2=1+1x3−2.
Again, as $x\rightarrow\infty$x→∞, the quantity $\frac{1}{x^3-2}\rightarrow0$1x3−2→0 and so the curve asymptotically approaches the horizontal line $y=1$y=1 from above. Similarly, as $x\rightarrow-\infty$x→−∞ the curve approaches $y=1$y=1 from below.
The vertical asymptote is found by solving $x^3-2=0$x3−2=0, the solution of which $x=\sqrt[3]{2}$x=3√2. The curve approaches the asymptote on each side moving upwards on one side and downwards on the other.
Here is the graph:
Most of the features predicted are shown on the graph. However, this particular graph shows another feature in the area around near the origin that was not predicted in either form of the function. At $x=0$x=0 it appears that the slope of the curve becomes $0$0. In fact, using calculus tools we can verify that this is indeed the case.
This example serves to remind us that while expressing functions in different forms is illuminating, they may not reveal all of its critical features.
Practice Questions
Question 1
Consider the function $f\left(x\right)=\frac{3x^2+8}{x^2}$f(x)=3x2+8x2.
Rewrite $f\left(x\right)$f(x) in the form $f\left(x\right)=k+\frac{a}{x^2}$f(x)=k+ax2.
Question 2
Consider the function $f\left(x\right)=\frac{x+9}{x-5}$f(x)=x+9x−5.
Rewrite $f\left(x\right)$f(x) in the form $f\left(x\right)=\frac{a}{x-h}+k$f(x)=ax−h+k.
Question 3
Consider the function $f\left(x\right)=\frac{x^2+4x+2}{\left(x+2\right)^2}$f(x)=x2+4x+2(x+2)2.
Rewrite $f\left(x\right)$f(x) in the form $f\left(x\right)=k-\frac{a}{\left(x+h\right)^2}$f(x)=k−a(x+h)2.