Lesson

Recall that rational functions are given by $y=\frac{P\left(x\right)}{Q\left(x\right)}$`y`=`P`(`x`)`Q`(`x`),$Q\left(x\right)\ne0$`Q`(`x`)≠0 where $P\left(x\right)$`P`(`x`) and $Q\left(x\right)$`Q`(`x`) are polynomials.

While computer software programs are now available to plot them, general methods for sketching certain classes of them have been developed over time, and it is these methods that we focus on here.

The first strategy to discuss is the addition of ordinates method and we will illustrate with a simple example.

Suppose we wish to draw the rational function given by $y=\frac{x^2+6}{2x}$`y`=`x`2+62`x`.

By division we can show that the function can be re-expressed as $y=\frac{1}{2}x+\frac{3}{x}$`y`=12`x`+3`x`, where we notice that the rational function is the sum of two separate functions - a linear function and a hyperbolic function.

If we sketch each of the separate functions and then, plot the ordinate sums, we will create the outline of the curve we seek. This is graphically shown here:

The blue and green rectangles show how the ordinates add together in this way.

The method however has limited value - divisions become too cumbersome and the complexity of the addition of ordinates can be a challenging process.

Certain factored forms of rational functions can be quickly sketched by understanding what a curve does around its roots and points of discontinuity.

Once understood the following strategy is an extremely powerful one. It is worth spending a little time going through the explanation. For the function $y=\frac{P\left(x\right)}{Q\left(x\right)}$`y`=`P`(`x`)`Q`(`x`) we consider $P\left(x\right)$`P`(`x`) and $Q\left(x\right)$`Q`(`x`) separately.

If $P\left(x\right)$`P`(`x`) can be expressed as a product of multiples of linear factors, such as $\left(x-1\right)\left(x+2\right)^2$(`x`−1)(`x`+2)2 or $\left(2x-1\right)^3$(2`x`−1)3 or $\left(1-x\right)\left(x-2\right)\left(x+5\right)$(1−`x`)(`x`−2)(`x`+5) etc, then the behaviour of the function around the corresponding roots is governed by the multiplicity of those roots.

The multiplicity of a root is determined by the number of times the associated factor is repeated. So for example a root at say $x=a$`x`=`a` with multiplicity $3$3 corresponds to the repeated factor $\left(x-a\right)^3$(`x`−`a`)3 in $P\left(x\right)$`P`(`x`).

Three multiplicities interest us:

Firstly, curves of functions like $y=\frac{1-x}{x+4}$`y`=1−`x``x`+4 or $y=\frac{x-1}{x+4}$`y`=`x`−1`x`+4, where a root of multiplicity $1$1 exists at $x=1$`x`=1 will cross through like a straight line through the root, as shown here.

Secondly, a root of multiplicity $2$2, with a repeated factor in $P\left(x\right)$`P`(`x`), will cause the curve to become tangential at the root, just like a parabola around its vertex does. The diagram here shows what would happen if $P\left(x\right)$`P`(`x`) contained a factor of either $\left(x+1\right)^2$(`x`+1)2 or $-\left(x+1\right)^2$−(`x`+1)2 .

Finally, a root of multiplicity $3$3 at causes the curve to form the shape below (known as a horizontal inflection). Here $P\left(x\right)$`P`(`x`) would contain the factor $\pm\left(x-3\right)^3$±(`x`−3)3.

If $Q\left(x\right)$`Q`(`x`) can be likewise factored into linear factors, the multiplicities of those factors determine the way the curve behaves around the discontinuity.

If the multiplicity is an *even* number, then the curve on both sides of the discontinuity rises (or falls) in *the same direction*. If the multiplicity is *odd*, the curve rises (or falls) in *opposite directions*.

The dotted lines in the diagram are often referred to as poles. Using this terminology, an even multiplicity corresponds to an *even pole* and an odd multiplicity corresponds to an *odd pole*.

Consider the following examples where we put the overall strategy into practice.

Sketch the rational function given by $y=\frac{x-2}{x\left(x-4\right)}$`y`=`x`−2`x`(`x`−4).

It has a single root of multiplicity $1$1, at $x=2$`x`=2. It also has two odd poles at $x=0$`x`=0 and $x=4$`x`=4. Here is the sketch:

The single root and the pole positions have been highlighted.

Note also the behaviour as $x\rightarrow\pm\infty$`x`→±∞. Because the degree of $P\left(x\right)$`P`(`x`) is less than the degree of $Q\left(x\right)$`Q`(`x`), the function approaches $0$0 (and thus the curve approaches the asymptote $y=0$`y`=0 ) as $x$`x` becomes more positively large or more negatively large.

**A sketching strategy**

In sketching this function, we might first pick any suitable point, say the point with $x=5$`x`=5, and $y=\frac{5-2}{5\left(5-4\right)}=\frac{3}{5}$`y`=5−25(5−4)=35 as highlighted in red in the diagram. This happens to locate the point to the right of the pole at $x=4$`x`=4 (Think of it as a point in the far right region of the three regions created by the two poles).

Think about how you would sketch the curve from this point. In drawing the curve leftward, you would have to negotiate the odd pole at $x=4$`x`=4. Drawing to the right the curve would asymptotically approach the $x$`x` axis. Next, the region between the poles must be drawn to reflect the oddness of both poles and the linear crossing at $x=2$`x`=2. The curve in the far left region must start from below the $x$`x` axis (part of an odd pole), but must not cross the x axis (there is only one root). It needs to bend and become asymptotic to the $x$`x` axis.

Sketch the rational function given by $y=\frac{1}{x^2}$`y`=1`x`2.

Because the numerator is the constant $1$1, the function has no roots (it cannot cross the $x$`x` axis). In the numerator we see an even pole at $x=0$`x`=0.

As $x\rightarrow\pm\infty$`x`→±∞, the curve asymptotically approaches the $x$`x` axis.

The point $\left(1,1\right)$(1,1) verifies that the curve is above the $x$`x`-axis.

Thus the sketch is straightforward:

Sketch the rational function given by $y=\frac{\left(x-2\right)^3}{x^2-25}$`y`=(`x`−2)3`x`2−25.

We see a root at $x=2$`x`=2 of multiplicity $3$3, and, because the denominator factorises to $\left(x+5\right)\left(x-5\right)$(`x`+5)(`x`−5), we know that there are two odd poles at $x=\pm5$`x`=±5.

The degree of the numerator is one higher than that of the denominator, so there we expect an oblique linear asymptote (The exact equation of the asymptote is difficult to determine, but it is accessible by division and is given by the line $y=x-6$`y`=`x`−6 ).

Here is the sketch:

The following applet allows you to create your own rational function. The equation form for this applet is given by:

$y=\frac{\left(x-a\right)^m\left(x-b\right)^n}{\left(x-c\right)^p\left(x-d\right)^q}$`y`=(`x`−`a`)`m`(`x`−`b`)`n`(`x`−`c`)`p`(`x`−`d`)`q`

where:

- the indices $m$
`m`, $n$`n`, $p$`p`and $q$`q`can be chosen as either $0,1,2$0,1,2 or $3$3 - the constants $a$
`a`, $b$`b`$c$`c`and $d$`d`can be chosen as integers in the range $-5\le x\le5$−5≤`x`≤5

Note that if you choose $a=c$`a`=`c`, $a=d$`a`=`d`, $b=c$`b`=`c` or $b=d$`b`=`d`, there will be common factors in the numerator and denominator of the rational function and this is likely to result in a less complex shape of the curve.

For example you could create functions like:

- $y=\frac{\left(x-3\right)\left(x+3\right)}{x}$
`y`=(`x`−3)(`x`+3)`x` - $y=\frac{\left(x+3\right)}{x^2}$
`y`=(`x`+3)`x`2 - $\frac{1}{\left(x-1\right)^3}$1(
`x`−1)3 - $y=\frac{\left(x-1\right)\left(x-3\right)^2}{\left(x+1\right)^2\left(x+5\right)}$
`y`=(`x`−1)(`x`−3)2(`x`+1)2(`x`+5)

Consider the graph of the rational function $f\left(x\right)$`f`(`x`).

Loading Graph...

How does the graph indicate that $f\left(-6\right)$

`f`(−6) and $f\left(6\right)$`f`(6) do not exist? Select all the correct options.There are no points on the graph at $x=6$

`x`=6 and $x=-6$`x`=−6AThere is a horizontal asymptote passing through $x=6$

`x`=6 and $x=-6$`x`=−6BThere are vertical asymptotes at $x=6$

`x`=6 and $x=-6$`x`=−6CThe graph of the function intercepts the vertical lines at $x=6$

`x`=6 and $x=-6$`x`=−6.DThere are no points on the graph at $x=6$

`x`=6 and $x=-6$`x`=−6AThere is a horizontal asymptote passing through $x=6$

`x`=6 and $x=-6$`x`=−6BThere are vertical asymptotes at $x=6$

`x`=6 and $x=-6$`x`=−6CThe graph of the function intercepts the vertical lines at $x=6$

`x`=6 and $x=-6$`x`=−6.DBy filling in the gaps, determine the domain of $f\left(x\right)$

`f`(`x`).Domain of $f$

`f`: $\left(-\infty,\editable{}\right)$(−∞,) $\cup$∪ $\left(\editable{},\editable{}\right)$(,) $\cup$∪ $\left(\editable{},\infty\right)$(,∞)Which of the following is the range of $f\left(x\right)$

`f`(`x`)?The range is all real numbers

A$\left(-\infty,-6\right)$(−∞,−6)$\cup$∪$0$0$\cup$∪$\left(6,\infty\right)$(6,∞)

B$\left(-\infty,0\right)$(−∞,0)$\cup$∪$\left(0,\infty\right)$(0,∞)

CThe range is all real numbers

A$\left(-\infty,-6\right)$(−∞,−6)$\cup$∪$0$0$\cup$∪$\left(6,\infty\right)$(6,∞)

B$\left(-\infty,0\right)$(−∞,0)$\cup$∪$\left(0,\infty\right)$(0,∞)

C

Consider the function $f\left(x\right)=\frac{x^2-4}{x-2}$`f`(`x`)=`x`2−4`x`−2.

What is the domain of the function?

$($($-\infty$−∞$,$,$\infty$∞$)$)

A$($($-\infty$−∞$,$,$4$4$)$)$\cup$∪$($($4$4$,$,$\infty$∞$)$)

B$($($-\infty$−∞$,$,$2$2$)$)$\cup$∪$($($2$2$,$,$\infty$∞$)$)

C$($($-\infty$−∞$,$,$-2$−2$)$)$\cup$∪$($($-2$−2$,$,$\infty$∞$)$)

D$($($-\infty$−∞$,$,$\infty$∞$)$)

A$($($-\infty$−∞$,$,$4$4$)$)$\cup$∪$($($4$4$,$,$\infty$∞$)$)

B$($($-\infty$−∞$,$,$2$2$)$)$\cup$∪$($($2$2$,$,$\infty$∞$)$)

C$($($-\infty$−∞$,$,$-2$−2$)$)$\cup$∪$($($-2$−2$,$,$\infty$∞$)$)

DWhat is the $y$

`y`value of the $y$`y`-intercept?What is the $x$

`x`value of the $x$`x`-intercept? Write each line of working as an equation.Which of the following functions $g\left(x\right)$

`g`(`x`) is equivalent to $f\left(x\right)$`f`(`x`)?$g\left(x\right)=x-2,x\ne-2$

`g`(`x`)=`x`−2,`x`≠−2A$g\left(x\right)=x^2-4$

`g`(`x`)=`x`2−4B$g\left(x\right)=x+2,x\ne2$

`g`(`x`)=`x`+2,`x`≠2C$g\left(x\right)=x+2$

`g`(`x`)=`x`+2D$g\left(x\right)=x-2,x\ne-2$

`g`(`x`)=`x`−2,`x`≠−2A$g\left(x\right)=x^2-4$

`g`(`x`)=`x`2−4B$g\left(x\right)=x+2,x\ne2$

`g`(`x`)=`x`+2,`x`≠2C$g\left(x\right)=x+2$

`g`(`x`)=`x`+2DGraph $f\left(x\right)$

`f`(`x`), making sure to indicate any holes:Loading Graph...

Consider the function $f\left(x\right)=\frac{x^2-16}{x}$`f`(`x`)=`x`2−16`x`

Write the function in the form $f\left(x\right)=\text{quotient }+\frac{\text{remainder }}{\text{divisor }}$

`f`(`x`)=quotient +remainder divisor .Find the equation of the oblique asymptote.

Plot both asymptotes on the graph below.

Loading Graph...Which of the following is the plot of $f\left(x\right)=\frac{x^2-16}{x}$

`f`(`x`)=`x`2−16`x`?Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

Consider the function $f\left(x\right)=\frac{x^5+3x^3+6x^2+11}{x^2+3}$`f`(`x`)=`x`5+3`x`3+6`x`2+11`x`2+3.

Rewrite the function in the form $P\left(x\right)+\frac{A}{x^2+3}$

`P`(`x`)+`A``x`2+3, where $P\left(x\right)$`P`(`x`) is a polynomial and $A$`A`is a constant.What is the equation of the non-linear asymptote of this function?

Form and use trigonometric, polynomial, and other non-linear equations

Apply trigonometric methods in solving problems