Linear Rational Functions (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Basic rational function

Find the key features and sketch the graph of $y = \frac{x+1}{x-4}$

Solution:

Step 1: Find the intercepts

When $x = 0$: $y = \frac{0+1}{0-4} = \frac{1}{-4} = -\frac{1}{4}$

So the y-intercept is (0, -1/4)

When $y = 0$: $0 = \frac{x+1}{x-4}$

The numerator must equal zero, so $x + 1 = 0$, giving $x = -1$

The x-intercept is (-1, 0)

Step 2: Find the vertical asymptote

Set the denominator equal to zero: $x - 4 = 0$ $x = 4$

The vertical asymptote is the line x = 4

Step 3: Find the horizontal asymptote

Rearrange to make $x$ the subject:

• $y = \frac{x+1}{x-4}$
• $y(x-4) = x+1$
• $xy - 4y = x + 1$
• $xy - x = 4y + 1$
• $x(y-1) = 4y + 1$
• $x = \frac{4y + 1}{y - 1}$

Set the denominator equal to zero:

• $y - 1 = 0$
• $y = 1$

The horizontal asymptote is the line y = 1

Step 4: Test behavior near asymptotes

When $x = 1000$ (large positive value):

$y = \frac{1000 + 1}{1000 - 4} = \frac{1001}{996} \approx 1.005$

The curve approaches $y = 1$ from above as $x \to +\infty$

When $x = -1000$ (large negative value):

$y = \frac{-1000 + 1}{-1000 - 4} = \frac{-999}{-1004} \approx 0.995$

The curve approaches $y = 1$ from below as $x \to -\infty$

Step 5: Sketch the curve

The curve has two branches: one passing through $(-1, 0)$ and $(0, -\frac{1}{4})$ on the left side of the vertical asymptote, approaching $y = 1$ from below as $x$ decreases. The other branch is on the right side of $x = 4$, approaching $y = 1$ from above as $x$ increases.

Worked Example 2: Function with algebraic constant

Find the key features and sketch the graph of $y = \frac{x+a}{x-2}$ where $a > 0$

Solution:

Intercepts:

When $x = 0$: $y = \frac{0+a}{0-2} = -\frac{a}{2}$

y-intercept: (0, -a/2)

When $y = 0$:

• $0 = \frac{x+a}{x-2}$
• $x + a = 0$
• $x = -a$

x-intercept: (-a, 0)

Vertical asymptote:

Set denominator to zero:

• $x - 2 = 0$
• $x = 2$

Horizontal asymptote:

As $x$ becomes very large:

$y = \frac{x+a}{x-2} \approx \frac{x}{x} = 1$

When $x = 1000$:

$y = \frac{1000 + a}{1000 - 2} = \frac{1000 + a}{998}$

Since $a > 0$, this gives $y > 1$, so the curve approaches from above.

When $x = -1000$:

$y = \frac{-1000 + a}{-1000 - 2} = \frac{-1000 + a}{-1002}$

Since $a$ is positive but much smaller than 1000, this gives $y < 1$, so the curve approaches from below.

Horizontal asymptote: y = 1

The sketch shows a hyperbola with the intercepts labeled in terms of $a$, with the vertical asymptote at $x = 2$ and horizontal asymptote at $y = 1$.

Worked Example 3: Finding intersections with a linear function

a) Identify the key features and sketch the graphs of $y = \frac{x}{x-5}$ and $y = 5x - 24$

b) Find the points of intersection

Solution:

Part a:

For $y = \frac{x}{x-5}$:

When $x = 0$: $y = 0$ (both intercepts at origin)

When $y = 0$: $x = 0$

Vertical asymptote: $x = 5$ (denominator zero)

Horizontal asymptote: Rearranging $y(x-5) = x$ gives $xy - 5y = x$, so $x(y-1) = 5y$, thus $x = \frac{5y}{y-1}$

Setting denominator to zero: y = 1

For the linear function $y = 5x - 24$:

• Gradient: $5$
• y-intercept: $-24$

Part b: Find intersections

Set the two functions equal:

$\frac{x}{x-5} = 5x - 24$

Multiply both sides by $(x-5)$:

• $x = (5x - 24)(x - 5)$
• $x = 5x^2 - 25x - 24x + 120$
• $x = 5x^2 - 49x + 120$
• $0 = 5x^2 - 50x + 120$
• $0 = x^2 - 10x + 24$

Factorise:

$0 = (x - 4)(x - 6)$

So $x = 4$ or $x = 6$

When $x = 4$: $y = 5(4) - 24 = -4$

When $x = 6$: $y = 5(6) - 24 = 6$

The points of intersection are (4, -4) and (6, 6)

Worked Example 4: Solving a rational inequality

Solve the inequality $\frac{18x - 135}{x - 3} > 21 - x$ graphically

Solution:

Step 1: Investigate key features

For $y = \frac{18x - 135}{x - 3}$:

• Vertical asymptote: $x = 3$
• When $y = 0$: $18x - 135 = 0$, so $x = 7.5$ (x-intercept: $(7.5, 0)$)
• When $x = 0$: $y = \frac{-135}{-3} = 45$ (y-intercept: $(0, 45)$)
• Horizontal asymptote: As $x \to \infty$, $y \approx \frac{18x}{x} = 18$

For $y = 21 - x$:

• This is a straight line with gradient $-1$ and y-intercept $21$

Step 2: Sketch and find intersections

Setting the functions equal:

• $\frac{18x - 135}{x - 3} = 21 - x$
• $18x - 135 = (21 - x)(x - 3)$
• $18x - 135 = 21x - 63 - x^2 + 3x$
• $18x - 135 = -x^2 + 24x - 63$
• $x^2 - 6x - 72 = 0$
• $(x + 6)(x - 12) = 0$

So $x = -6$ or $x = 12$

When $x = -6$: $y = 21 - (-6) = 27$, giving point $(-6, 27)$

When $x = 12$: $y = 21 - 12 = 9$, giving point $(12, 9)$

Step 3: Determine solution

The inequality $\frac{18x - 135}{x - 3} > 21 - x$ is satisfied where the rational function (curve) is above the straight line.

From the graph, this occurs when: $-6 < x < 3 \text{ or } x > 12$

Note that $x = 3$ is excluded because it makes the denominator zero.

Worked Example 5: Comparing two rational functions

Following an experiment, student A's results fit the graph of $y(x - 3) = 2$, whilst student B's results fit $y(x + 2) = 3$

a) Find the point of intersection of their graphs

b) Find the range of values of $x$ where student A gives higher results than student B

Solution:

Part a:

Rewrite the equations: $y = \frac{2}{x - 3} \quad \text{and} \quad y = \frac{3}{x + 2}$

At the point of intersection: $\frac{2}{x - 3} = \frac{3}{x + 2}$

Cross-multiply: $2(x + 2) = 3(x - 3)$ $2x + 4 = 3x - 9$ $13 = x$

When $x = 13$: $y = \frac{2}{13 - 3} = \frac{2}{10} = 0.2$

The point of intersection is (13, 0.2)

Part b:

First, investigate the key features:

For $y = \frac{2}{x - 3}$:

• Asymptotes: $x = 3$ and $y = 0$

For $y = \frac{3}{x + 2}$:

• Asymptotes: $x = -2$ and $y = 0$

Student A's results are higher than student B's when the blue graph is above the red graph. From the sketch, this occurs when: $x < -2 \text{ or } 3 < x < 13$

Algebraic confirmation:

We need to solve $\frac{2}{x - 3} > \frac{3}{x + 2}$

Case 1: Both denominators positive or both negative

If both $(x - 3)$ and $(x + 2)$ have the same sign:

• $2(x + 2) > 3(x - 3)$
• $2x + 4 > 3x - 9$
• $13 > x$

Combined with the condition that $x > 3$ or $x < -2$, this gives: $x < -2 \text{ or } 3 < x < 13$

Case 2: One denominator positive and one negative

If $(x + 2)(x - 3) < 0$ (i.e., $-2 < x < 3$):

• $2(x + 2) < 3(x - 3)$
• $2x + 4 < 3x - 9$
• $13 < x$

But this contradicts $-2 < x < 3$, so there's no solution in this case.

Therefore, the solution is x < -2 or 3 < x < 13, confirming our graphical result.