Rational Functions with Oblique Asymptotes (AQA A-Level Further Maths): Revision Notes

Example: Proper Rational Function

Consider the function $y = \frac{x}{x^2 - 4}$

This function has:

• Vertical asymptotes at $x = 2$ and $x = -2$ (where the denominator equals zero)
• Horizontal asymptote at $y = 0$ (because the denominator degree is greater)

As $x$ gets very large, the $x^2$ term in the denominator grows much faster than the $x$ in the numerator, causing the function to approach zero.

Example: Improper Rational Function

Consider the function $y = \frac{x^2}{x - 4}$

This function has:

• Vertical asymptote at $x = 4$ (where the denominator equals zero)
• Oblique asymptote (shown in the graph below)

Worked Example: Finding Asymptotes

Question: Find the equations of the asymptotes of the curve $y = \frac{x^2 - x}{x + 1}$

Solution:

Step 1: Find the vertical asymptote.

There is a vertical asymptote at $x = -1$ because the function is undefined when $x = -1$ (the denominator would be zero).

Step 2: Use algebraic long division to divide out the function.

Dividing $x^2 - x$ by $x + 1$:

Performing the division:

• $x^2 \div x = x$
• Multiply: $x(x + 1) = x^2 + x$
• Subtract: $(x^2 - x) - (x^2 + x) = -2x$
• Bring down: $-2x$
• $-2x \div x = -2$
• Multiply: $-2(x + 1) = -2x - 2$
• Subtract: $-2x - (-2x - 2) = 2$

Therefore: $y = x - 2 + \frac{2}{x + 1}$

Step 3: Identify the oblique asymptote.

As $x \to \pm\infty$, the term $\frac{2}{x+1} \to 0$

Hence $y \to x - 2$

There is an oblique asymptote with equation $y = x - 2$

Worked Example: Finding Unknowns Given an Asymptote

Question: A curve with equation $y = \frac{ax^2 + bx + 1}{x - 2}$ has an asymptote $y = 2x + 1$

a) Find the values of $a$ and $b$

b) Write down the equation of the other asymptote

c) Without using calculus, find the coordinates of the turning points

d) Sketch the curve

Solution:

Part a) Find the values of $a$ and $b$

The oblique asymptote is found by dividing the numerator by the denominator. If the asymptote is $y = 2x + 1$, then when we multiply this by the denominator, we should get the numerator (plus a remainder).

Multiply the asymptote by the denominator: $(2x + 1)(x - 2) = 2x^2 - 4x + x - 2 = 2x^2 - 3x - 2$

Comparing with $ax^2 + bx + 1$:

The coefficient of $x^2$ gives: $a = 2$

The coefficient of $x$ gives: $b = -3$

The rational function is $y = \frac{2x^2 - 3x + 1}{x - 2}$

Part b) Write down the equation of the other asymptote

The vertical asymptote occurs where the denominator equals zero:

$x - 2 = 0$

Therefore: $x = 2$

Part c) Find the coordinates of the turning points

The range of allowable values of $y$ can be found by setting $y = k$ and solving for $k$:

For real solutions, the discriminant must be non-negative:

Using the quadratic formula to find when $k^2 - 10k + 1 = 0$:

$k = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2} = 5 \pm 2\sqrt{6}$

Therefore: $k \geq 5 + 2\sqrt{6}$ or $k \leq 5 - 2\sqrt{6}$

The curve exists for $y \geq 5 + 2\sqrt{6}$ and $y \leq 5 - 2\sqrt{6}$

There will be a minimum point at $y = 5 + 2\sqrt{6}$ and a maximum point at $y = 5 - 2\sqrt{6}$

To find the $x$-values for these turning points, substitute the $y$-values into the original function:

For the minimum: $(5 + 2\sqrt{6})(x - 2) = 2x^2 - 3x + 1$

Solving: $x = \frac{4 + \sqrt{6}}{2}$

For the maximum: $(5 - 2\sqrt{6})(x - 2) = 2x^2 - 3x + 1$

Solving: $x = \frac{4 - \sqrt{6}}{2}$

Turning points:

• Minimum at $\left(\frac{4 + \sqrt{6}}{2}, 5 + 2\sqrt{6}\right)$
• Maximum at $\left(\frac{4 - \sqrt{6}}{2}, 5 - 2\sqrt{6}\right)$

Part d) Sketch the curve

Draw both asymptotes (the vertical line $x = 2$ and the oblique line $y = 2x + 1$). Mark the turning points and sketch the curve approaching the asymptotes.