Given enough information, we can find the equation of any rational function of the form $y=\frac{P\left(x\right)}{Q\left(x\right)}$y=P(x)Q(x). We discuss two examples.
Example 1
Suppose, for example, we were shown this graph of a rational function, and were asked to determine its equation.
We notice the four features as marked on the graph.
It has one root of multiplicity $2$2 at $x=0$x=0, and therefore the numerator function must be given by $P\left(x\right)=kx^2$P(x)=kx2 for some non-zero constant $k$k.
It has a discontinuity at $x=3$x=3, with the curve approaching the 'pole' asymptotically in different directions. It also has an oblique linear asymptote, meaning that the degree of $Q\left(x\right)$Q(x) is one less than the degree of $P\left(x\right)$P(x). These two facts together imply that that the denominator function $Q\left(x\right)=\left(x-3\right)$Q(x)=(x−3).
Hence the rational function is of the form $y=\frac{kx^2}{x-3}$y=kx2x−3, and we need only substitute the point $\left(6,12\right)$(6,12) to reveal $k$k.
| $y$y | $=$= | $\frac{kx^2}{x-3}$kx2x−3 |
| $12$12 | $=$= | $\frac{k\left(6^2\right)}{6-3}$k(62)6−3 |
| $12$12 | $=$= | $\frac{36k}{3}$36k3 |
| $12$12 | $=$= | $12k$12k |
| $\therefore$∴ $k$k | $=$= | $1$1 |
Hence our function is given by $y=\frac{x^2}{x-3}$y=x2x−3. Note that if we divide $P\left(x\right)=x^2$P(x)=x2 by $Q\left(x\right)=x-3$Q(x)=x−3 we find that the rational function can be expressed as $y=\left(x+3\right)+\frac{9}{x-3}$y=(x+3)+9x−3. This means that as $x\rightarrow\infty$x→∞, the quantity $\frac{9}{x-3}\rightarrow0$9x−3→0 and so the oblique asymptote has the equation $y=\left(x+3\right)$y=(x+3), and this is clearly shown in the graph.
Example 2
Determine the equation of the graph depicted here:
The only root is clearly of multiplicity $1$1, and so $P\left(x\right)=kx$P(x)=kx.
The two odd poles at $x=-4$x=−4 and $x=4$x=4 divide the curve into three sections. This implies that the denominator function $Q\left(x\right)=\left(x-4\right)\left(x+4\right)=x^2-16$Q(x)=(x−4)(x+4)=x2−16.
Note that we can rule out $Q\left(x\right)$Q(x) being $\left(x-4\right)^3\left(x+4\right)^3$(x−4)3(x+4)3 on the basis that the higher powers would critically reduce the size of the curve's ordinates and it doesn't appear from the graph that this is the case.
In addition, we do not have to concern ourselves with the constant in from of the function for this is already covered by the constant $k$k in $P\left(x\right)$P(x).
So we conclude that the rational function has the form $y=\frac{kx}{x^2-16}$y=kxx2−16.
The point shown in the diagram is $\left(8,2\right)$(8,2), so using this information, we have:
| $2$2 | $=$= | $\frac{8k}{8^2-16}$8k82−16 |
| $2$2 | $=$= | $\frac{8k}{48}$8k48 |
| $8k$8k | $=$= | $96$96 |
| $\therefore$∴ $k$k | $=$= | $12$12 |
Hence the rational function is given by $y=\frac{12x}{x^2-16}$y=12xx2−16.
Also note that as $x\pm\rightarrow\infty,y\rightarrow0$x±→∞,y→0 and this behaviour is observed in the graph.