Further Pure | Topic 1: Polynomials and Rational Functions(Part 2)

Curriculum Objective:

• Sketch graphs of simple rational functions, including the determination of oblique asymptotes, in cases where the degree of the denominator and numerator are no more than 2 (detailed plotting of the curve will not be required, but sketches will generally be expected to show significant features)

Curve sketching:

As stated above the curve sketching would require a certain amount of detail. The intersection of the curve with either axes, the turning point of the curve and asymptotes.

Let us begin by defining asymptotes. An asymptote is a straight line that becomes a tangent to the curve as either $x \to  \pm \infty $ or $y \to  \pm \infty $.

To understand what that means consider the curve $y$:

$y = \frac{{x + 1}}{{x - 3}}$

For this curve the point x = 3 is undefined, for then we will have a zero as the denominator, but consider what might happen as we take x to be close to 3, say 3.0001 or 3.00000001. y would begin to become greater and greater still, a trend that may be succinctly described in mathematical notation as $y \to \infty $ or as y approaches infinity or positive infinity. What if we took the balue 2.9999 or 2.999999999999. In that case y would approach negative infinity. It follows therefore, the line $x = 3$ is vertical asymptote.

Having considered this, the logical consequence would be the question what would happen, if say x approached either negative or positive infinity. To examine that the function will have to be manipulated somewhat, into a form more commensurate with our method of analysis. If you were to divided $x + 1$ by $x - 3$ we get: 

$y = 1 + \frac{4}{{x - 3}}$

Now consider what might happen if x approaches either positive or negative infinity. The fraction would become infinitely small, if you will, to point that it becomes completely negligible. A fact represted mathematically as

$\mathop {\lim }\limits_{x \to \infty } \left( {\frac{4}{{x - 3}}} \right) = 0$

While what I have written involves some fairly beautiful mathematics, you may only take it to mean that that fraction becomes zero as x become infinitely large and the original curve would then reduce to $y = 1$. This line $y = 1$ is then the horizontal asymptote.

We shall consider oblique asymptotes in the next example, for now let us plot this curve. First we find the points of intersection with axes of the curves. These are (-1,0) and (0,(-1/3)). Next we must find the turning point(s), of which there are none.

The sketch with this information should look like this. You should take note of the fact that the sketch referenced doesn't indicate the asymptotes in your sketches however you should make them apparent.

We now solve Question 11 either, from the 2013 October/November Session Paper 12.

$y = \frac{{p{x^2} + 4x + 1}}{{x + 1}}$

i) Obtain the equations of the asymptote of the curve

ii) Find the value of p for which the x-axis is a tangent to the curve, and sketch the curve in this case

iii) For the case p = 1, show that the curve has no turning points, and sketch the curve in this case.

Sol:

i) The vertical asymptote is obviously x = -1

For the oblique asymptote we again carry out the division to bring the function into the form:

$y = px + 4 - p + \frac{{p - 3}}{{x + 1}}$

Wherefrom we can glean that the oblique asymptote is $y = px + 4 - p$

ii) At the x-axis:

 $\frac{{p{x^2} + 4x + 1}}{{x + 1}} = 0$
$p{x^2} + 4x + 1 = 0$

For the x-axis to be a tangent to the curve, I would expect the discriminant $b^2 - 4ac = 0$. Therefore, inputting the values:

${(4)^2} - 4(p)(1) = 0$
$4p = 16$
$p = 4$

For this value of p, the vertical asymptote is evidently unchanged, the oblique asymptote will be:

$y = 4x$

For the y-intercept we have ($ - \frac{1}{2}$, 0), for the x-intercept, we find the point (1, 0). Next we consider the turning points which are ($ - \frac{3}{2},8$) and ($ - \frac{1}{2},0$)

The plot with this information is this.

You should familiarise yourself with the general form of such sketches. Another deduction that can be made from the equation of the curve, that you may find helpful in your sketching is the side of the asymptote that the curves lies on. In the previous case we have the equation:

$y = 4x + \frac{{1}}{{x + 1}}$

Now consider that for the values of x > -1, the fraction $\frac{1}{{x + 1}}$ will be positive and therefore the curve, for these values of x, will be greater than than, or will lie above the oblique asymptote. A similar argument suggests that for values of x less than -1, the curve will lie below the oblique asymptote.

iii) This is left to the reader.