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

Worked Example 1: Finding Intercepts and Asymptotes

Question: For the graph of $y = \frac{x^2 - 16}{x^2 + 4}$, find the intercepts where the curve crosses the axes, and show that there is no vertical asymptote but that there is a horizontal asymptote at $y = 1$.

Solution:

Finding intercepts:

When $x = 0$: $y = \frac{-16}{4} = -4$

So the curve crosses the y-axis at (0, -4).

When $y = 0$:

$x^2 - 16 = 0$

This requires the numerator to equal zero (provided $x^2 + 4 \neq 0$, which is always true).

• $x^2 - 16 = 0$
• $x^2 = 16$
• $x = \pm 4$

So the curve crosses the x-axis at (4, 0) and (-4, 0).

Finding asymptotes:

For vertical asymptotes, we need $x^2 + 4 = 0$, which gives $x^2 = -4$. This has no real solutions, so there is no vertical asymptote.

For the horizontal asymptote, consider the behavior as x becomes very large:

$\text{As } x \to \infty, \quad \frac{x^2}{x^2} = 1$

So y = 1 is a horizontal asymptote.

Alternative verification: Substitute $x = \pm 1000$ to show the graph approaches the asymptote from above on both sides of the graph.

To find where the curve crosses this asymptote, solve $y = 1$:

• $\frac{x^2 - 16}{x^2 + 4} = 1$
• $x^2 - 16 = x^2 + 4$
• $-16 = 4$

This is impossible, so the curve never actually crosses the horizontal asymptote. Therefore, the range is restricted: $y < 1$ (since the curve approaches 1 from below).

Worked Example 2: Finding Range and Turning Points

Question: Find the range of x and y values for which the graph of $y = \frac{x^2 + 6x + 9}{x^2 + 3x + 3}$ exists. Use this information to find any turning points.

Solution:

Finding x-values:

The graph exists for all values of x except where there are vertical asymptotes. Vertical asymptotes occur when the denominator equals zero.

Solve $x^2 + 3x + 3 = 0$:

Using the discriminant: $b^2 - 4ac = 9 - 12 = -3 < 0$

Since the discriminant is negative, there are no real solutions. This means the graph exists for all real values of x.

Finding y-values:

To find the range of y, consider where the graph crosses the line $y = k$ and deduce allowable values for k.

Let $y = k$ and rearrange:

• $k(x^2 + 3x + 3) = x^2 + 6x + 9$
• $kx^2 + 3kx + 3k = x^2 + 6x + 9$
• $(k-1)x^2 + 3(k-2)x + 3(k-3) = 0$

For x to take real values when $b^2 - 4ac \geq 0$:

• $[3(k-2)]^2 - 4(k-1) \times 3(k-3) \geq 0$
• $9(k-2)^2 - 12(k-1)(k-3) \geq 0$
• $9(k^2 - 4k + 4) - 12(k^2 - 4k + 3) \geq 0$
• $9k^2 - 36k + 36 - 12k^2 + 48k - 36 \geq 0$
• $-3k^2 + 12k \geq 0$
• $3k(k - 4) \leq 0$

This inequality holds when 0 ≤ k ≤ 4.

Therefore, the curve only exists for 0 ≤ y ≤ 4.

Finding turning points:

As x gets very large, $y \to \frac{x^2}{x^2} = 1$, so $y = 1$ is a horizontal asymptote.

Since $y = 1$ is the only horizontal asymptote and $0 \leq y \leq 4$, the values $y = 0$ and $y = 4$ must correspond to either minimum or maximum points.

When $y = 4$:

• $3x^2 + 6x + 3 = 0$
• $3(x+1)^2 = 0$
• $x = -1$

So the maximum point is at (-1, 4).

When $y = 0$:

• $x^2 + 6x + 9 = 0$ (provided $x^2 + 3x + 3 \neq 0$)
• $(x+3)^2 = 0$
• $x = -3$

So the minimum point is at (-3, 0).

Worked Example 3: Complete Curve Analysis

Question: Identify the key features and sketch the graph of $y = \frac{x^2 - x - 6}{x^2 - x - 12}$.

Solution:

Finding intercepts:

When $x = 0$: $y = \frac{-6}{-12} = \frac{1}{2}$

So the y-intercept is at (0, 1/2).

When $y = 0$:

• $x^2 - x - 6 = 0$ (provided $x^2 - x - 12 \neq 0$)
• $(x-3)(x+2) = 0$
• $x = 3 \text{ or } x = -2$

Check: $(3)^2 - (3) - 12 = -6 \neq 0$ ✓ and $(-2)^2 - (-2) - 12 = -6 \neq 0$ ✓

So the curve crosses the axes at (0, 1/2), (-2, 0), and (3, 0).

Finding asymptotes:

For vertical asymptotes, solve $x^2 - x - 12 = 0$:

• $(x-4)(x+3) = 0$
• $x = 4 \text{ or } x = -3$

So there are vertical asymptotes at x = -3 and x = 4.

For the horizontal asymptote:

$\text{As } x \to \infty, \quad \frac{x^2}{x^2} \to 1$

So y = 1 is a horizontal asymptote. Substituting $x = \pm 1000$ confirms the graph approaches this asymptote from above on both sides.

Finding the range:

Let $y = k$ and rearrange:

• $k(x^2 - x - 12) = x^2 - x - 6$
• $kx^2 - kx - 12k = x^2 - x - 6$
• $(k-1)x^2 - (k-1)x - (12k-6) = 0$
• $(k-1)x^2 - (k-1)x + (6-12k) = 0$

For real x values, the discriminant must be non-negative:

• $(k-1)^2 - 4(k-1)(6-12k) \geq 0$
• $(k-1)^2 - 4(k-1)(6-12k) \geq 0$

Factoring out $(k-1)$:

• $(k-1)[(k-1) - 4(6-12k)] \geq 0$
• $(k-1)[k - 1 - 24 + 48k] \geq 0$
• $(k-1)[49k - 25] \geq 0$

This gives: $k \leq \frac{25}{49}$ or $k \geq 1$

However, because of the asymptote at $y = 1$, the function does not actually equal 1 except at specific points. The actual range is:

$y \leq \frac{25}{49} \text{ or } y \geq 1$

Finding turning points:

Since $y = 1$ is an asymptote as $x \to \pm \infty$, we only need to check $y = \frac{25}{49}$.

When $y = \frac{25}{49}$:

$\left(\frac{25}{49} - 1\right)x^2 + \left(1 - \frac{25}{49}\right)x + \left(6 - 12 \times \frac{25}{49}\right) = 0$

$\frac{-24}{49}x^2 + \frac{24}{49}x - \frac{6}{49} = 0$

• $-24x^2 + 24x - 6 = 0$
• $4x^2 - 4x + 1 = 0$
• $(2x-1)^2 = 0$
• $x = \frac{1}{2}$

So there is a turning point at (1/2, 25/49), which is a maximum point.

Worked Example 4: Curves Crossing Asymptotes

Question: Sketch the curve $y = \frac{x^2 + x - 6}{x^2 - x - 6}$.

Solution:

First, factorise both numerator and denominator: $y = \frac{(x+3)(x-2)}{(x-3)(x+2)}$

Critical values: The function has potential issues at x = -3, -2, 2, 3.

Sign analysis: Consider the sign of y between and on either side of these values:

• $x < -3$: y is negative
• $-3 < x < -2$: y is positive
• $-2 < x < 2$: y is negative
• $2 < x < 3$: y is positive
• $x > 3$: y is negative

Notice that every time a critical value of x is passed, y changes from one side of the x-axis to the other.

Intercepts: When $x = 0$, $y = 1$, so the curve passes through (0, 1).

Vertical asymptotes: At x = -2 and x = 3 (where the denominator equals zero).

Horizontal asymptote: As $x \to \infty$, $y \to \frac{x^2}{x^2} \to 1$, so y = 1 is a horizontal asymptote.

Crossing the horizontal asymptote: Solve $y = 1$:

• $\frac{x^2 + x - 6}{x^2 - x - 6} = 1$
• $x^2 + x - 6 = x^2 - x - 6$
• $2x = 0$
• $x = 0$

So the curve crosses the horizontal asymptote at (0, 1).

Although the function is not continuous (due to the vertical asymptotes), the curve actually passes through every value of y between the separate branches.