Reciprocal and Modulus Graphs (AQA A-Level Further Maths): Revision Notes

Worked Example: Quadratic Reciprocal Graph

Given: $f(x) = 2x^2 + x - 1$

a) Sketch the graph of $y = f(x)$

First, we identify the key features of this quadratic function:

• The y-intercept is at $(0, -1)$ (substitute $x = 0$)
• To find x-intercepts, set $f(x) = 0$: $2x^2 + x - 1 = 0$
• Factoring: $(2x - 1)(x + 1) = 0$
• x-intercepts are at $x = \frac{1}{2}$ and $x = -1$

To find the vertex, complete the square:

$y = 2\left(x + \frac{1}{4}\right)^2 - \frac{9}{8}$

The minimum point is at $\left(-\frac{1}{4}, -\frac{9}{8}\right)$.

b) On the same axes, sketch $y = \frac{1}{f(x)}$

Now we use the properties of reciprocal graphs to transform our sketch:

Step 1: Identify vertical asymptotes Since $f(x)$ has roots at $x = -1$ and $x = \frac{1}{2}$, the reciprocal function will have vertical asymptotes at these values. The function $\frac{1}{f(x)}$ is undefined at these points.

Step 2: Find intersection points We solve $f(x) = 1$:

$2x^2 + x - 1 = 1$

$2x^2 + x - 2 = 0$

$x = \frac{-1 \pm \sqrt{17}}{4} \approx 0.78 \text{ and } -1.28$

The graphs intersect at approximately $(0.78, 1)$ and $(-1.28, 1)$.

We also solve $f(x) = -1$:

$2x^2 + x - 1 = -1$

$2x^2 + x = 0$

$x(2x + 1) = 0$

$x = 0 \text{ or } x = -\frac{1}{2}$

The graphs intersect at $(0, -1)$ and $(-0.5, -1)$.

Step 3: Analyze turning points and gradient

The minimum point of $f(x)$ at $x = -0.25$ becomes a maximum point of $\frac{1}{f(x)}$ at the same x-coordinate.

Since $y = f(x)$ is increasing for $x > -0.25$ and decreasing for $x < -0.25$, the reciprocal $y = \frac{1}{f(x)}$ decreases for $x > -0.25$ and increases for $x < -0.25$.

Step 4: Determine signs

Both $f(x)$ and $\frac{1}{f(x)}$ are negative in the region $-1 < x < 0.5$, and positive elsewhere (excluding the asymptotes).

Worked Example: Reciprocal from a Given Graph

Given: A graph showing $y = f(x)$ with key points at $(-3, 0)$, $(0, 0)$, $(5, 0)$, a local maximum at $\left(-\frac{5}{3}, \frac{400}{27}\right)$, and a local minimum at $(3, -36)$.

Solution:

Step 1: Identify vertical asymptotes The function has three roots at $x = -3$, $x = 0$, and $x = 5$. Therefore, $y = \frac{1}{f(x)}$ will have three vertical asymptotes at these x-values.

Step 2: Transform turning points

• The maximum point of $f(x)$ becomes a minimum point of $\frac{1}{f(x)}$
• Since the maximum value is $\frac{400}{27}$, the minimum value of $\frac{1}{f(x)}$ is $\frac{27}{400}$
• The minimum point is at $\left(-\frac{5}{3}, \frac{27}{400}\right)$

Similarly:

• The minimum point of $f(x)$ at $(3, -36)$ becomes a maximum point of $\frac{1}{f(x)}$
• Since the minimum value is $-36$, the maximum value of $\frac{1}{f(x)}$ is $-\frac{1}{36}$
• The maximum point is at $\left(3, -\frac{1}{36}\right)$

Step 3: Analyze gradient behavior

• Between the maximum and minimum points of $f(x)$, the function is decreasing
• Therefore, $\frac{1}{f(x)}$ will increase in this region
• Elsewhere, where $f(x)$ is increasing, $\frac{1}{f(x)}$ will decrease

Step 4: Behavior near asymptotes

• When $f(x)$ approaches zero, $\frac{1}{f(x)}$ approaches infinity
• When $f(x)$ approaches infinity (becomes very large), $\frac{1}{f(x)}$ approaches zero
• The reciprocal function will have three vertical asymptotes, with the graph approaching $\pm \infty$ on either side depending on the sign of $f(x)$

Worked Example: Linear and Quadratic Modulus

a) Sketch $y = f(x)$ and $y = |f(x)|$ for $f(x) = 2x - 4$

The function $y = 2x - 4$ is a straight line with gradient 2.

• x-intercept: $(2, 0)$
• y-intercept: $(0, -4)$

For $y = |2x - 4|$, we reflect the negative part (where $x \< 2$) in the x-axis. The modulus graph crosses the axes at $(2, 0)$ and $(0, 4)$.

b) Sketch $y = f(x)$ and $y = |f(x)|$ for $f(x) = 2x^2 + 3x - 9$

This is a positive quadratic curve (U-shaped).

• x-intercepts: $(-3, 0)$ and $\left(\frac{3}{2}, 0\right)$
• y-intercept: $(0, -9)$

For $y = |2x^2 + 3x - 9|$, we reflect the negative part (between the roots) in the x-axis. The modulus graph crosses the x-axis at $(-3, 0)$ and $\left(\frac{3}{2}, 0\right)$, and has a y-intercept at $(0, 9)$.

Worked Example: Modulus of a General Curve

Given: A graph of $y = f(x)$ showing points $(-0.2, 0)$, $(1, 2.8)$, $(2.8, 0)$, $(4, -1.7)$, and $(5, 0)$.

Sketch $y = |f(x)|$

To create the modulus graph:

• Keep all parts of the graph that are above or on the x-axis
• Reflect the negative portion (around $x = 4$) in the x-axis
• The minimum point $(4, -1.7)$ becomes a maximum point $(4, 1.7)$

The red sections show the reflected portions, while the blue sections remain unchanged.

Worked Example: Combined Transformations and Inequalities

The sketch shows part of the graph of $f(x)$.

a) Find the coordinates of the images of the points $(-2, -12)$, $(0, -4)$ and $(2, 4)$ when $f(x)$ is transformed into $|f(x)|$

Under the transformation $f(x) \rightarrow |f(x)|$:

• $(-2, -12) \rightarrow$ $(-2, 12)$ — negative y-coordinate becomes positive
• $(0, -4) \rightarrow$ $(0, 4)$ — negative y-coordinate becomes positive
• $(2, 4) \rightarrow$ $(2, 4)$ — positive y-coordinate stays the same

All negative parts of the graph are reflected in the x-axis.

b) Sketch the graph of $|f(x)|$ and use it to solve $|f(x)| \leq 4$

We reflect all negative parts of the graph upward. The horizontal line $y = 4$ intersects $|f(x)|$ at $(0, 4)$ and $(2, 4)$.

Therefore, $|f(x)| \leq 4$ when $0 \leq x \leq 2$.

c) Sketch the graph of $\frac{1}{f(x)}$ and use it to estimate the solution to $\frac{1}{f(x)} \< 0$

For the reciprocal graph:

• The asymptote occurs approximately at $x = 1.6$ (where $f(x) = 0$)
• The point where $f(x) = 0$ becomes a vertical asymptote
• As $f(x)$ becomes very large (either positive or negative), $\frac{1}{f(x)}$ tends to zero

Since $\frac{1}{f(x)} < 0$ when $f(x) < 0$, and from the graph, $f(x) < 0$ for $x < 1.6$ (approximately), the solution is $x < 1.6$ (approximately).