Before reading this guide, you may find it helpful to read the guide on graphs from Module 7 (M7) and the guide on indices from Module 8 (M8).
A recap on graphs
In Module 7 (M7) Graphs, you learned that simultaneous equations, where one of the graphs is quadratic and the other is a straight line, can be solved by reading the points of intersection.
For Module 8 (M8), you need to
know more about using a quadratic graph to solve further equations by drawing an appropriate straight line
be able to recognise a further type of function and its graph – known as the exponential function
be able to interpret the gradient at any point on a curve
Using a quadratic graph to solve further equations
To use a quadratic graph to solve further equations:
- Start with the equation you have been asked to solve
- Find what needs to be added to the LHS of the new equation to give the equation of the graph already drawn.
- Draw this straight line
- Read off the \(x\) values of the points of intersection
Example
The graph of \(y = 2x^{2} + x – 6\) is shown below.
By drawing a suitable straight line on the same grid, find solutions for the equation \(2x^{2} + 2x – 8 = 0\)
Solution
- Start with the equation you have been asked to solve.
\(2x^{2} + 2x – 8 = 0\)
- Find what needs to be added to the left-hand side (LHS) of the new equation to give the equation of the graph already drawn.
\(2x^{2} + 2x – 8 = 0\) \(–x +2 \longrightarrow\) \(2x^{2} + x – 6\)
- Draw this straight line.
Draw the line \(y = –x + 2\) on the grid.
Question
Assume that the graph of \(y = x^{2} + x + 1\) has been drawn.
Find the equation of the line that needs to be drawn to solve the equation \(x^{2} – x – 3 = 0\).
There is no need to draw either of these graphs.
- Start with the equation you have been asked to solve.
\(x^{2} – x – 3 = 0\)
- Find what needs to be added to the left-hand side (LHS) of the new equation to give the equation of the graph already drawn.
\(x^{2} – x – 3 + \) \(2x + 4 \longrightarrow\) \(x^{2} + x + 1\)
The line to be drawn would be \(y = 2x + 4\).
Exponential graphs
Exponential graphs are graphs in the form \(y = k^{x}\). These graphs increase rapidly in the \(y\) direction and will never fall below the \(x\)-axis.
An exponential graph will look like this:
Example
Let's look at \(y = 2^{x}\).
\(x = –2, y = 2^{–2} = 0.25\)
\(x = –1, y = 2^{–1} = 0.5\), etc.
| \(x\) | −2 | −1 | 0 | 1 | 2 | 3 | 4 | 5 |
| \(y = 2^{x}\) | 0.25 | 0.5 | 1 | 2 | 4 | 8 | 16 | 32 |
Question
Complete the table for \(y = 2^{–x} (-3 \le x \le 3)\) and draw its graph.
| \(x\) | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
| \(2^{-x}\) |
You will need a grid like this:
First, complete the table, which will help you plot the points on the grid.
| \(x\) | −3 | −2 | −1 | 0 | 1 | 2 | 3 |
| \(2^{-x}\) | 8 | 4 | 2 | 1 | \(\frac{1}{2}\) | \(\frac{1}{4}\) | \(\frac{1}{8}\) |
Then plot the points and draw a line through them, which should curve like this:
Question
Use the graph of \(y = 2^{-x}\) to estimate the value of \(x\) for which \(2^{-x} = 6\).
Answer
Read the \(x\) value from the graph, where \(y = 6\).
When \(y = 6, x = –2.6\)
Gradient of a curve
The gradient of a curve at any point is a value which reflects the steepness of the curve at that point.
In the graph below, the curve is much steeper at point B than at point A.
The gradient of a curve at any point can be found by drawing a tangent to the curve and finding the gradient of the tangent.
Example
Part of the graph of \(y = 2x^{2}\) is shown below.
Find the gradient of the curve at the point where \(x = 2\).
Solution
Draw a tangent to the curve at the point where \(x = 2\).
Complete a right-angled triangle with the tangent as its hypotenuse.
\(\text{gradient of tangent} = \frac{\text{rise}}{\text{run}} = \frac{8}{1}\)
Gradient of curve when \(x = 2\) is \(8\)
Exponential growth and decay
Exponential graphs can be used to represent many real-life situations including growth of bacteria, compound interest, radioactive decay, and population increase.
Example
The graph shows how bacteria (\(y\)) grows over time (\(t\)). The equation of the graph is \(y = 20g^{t}\).
Determine the value of \(g\).
Solution
From the graph, when \(t = 2, y = 2000\).
Substituting these values into \(y = 20g^{t}\):
\(2000 = 20g^{2}\)
Dividing by 20 gives:
\(100 = g^2\)
Square rooting gives:
\(10 = g\)
Question
The half-life of a decaying radioactive substance is the time taken for the mass to be reduced by half.
The mass (\(M\)) of a particular radioactive substance \(x\) years after its production is shown on the graph below.
Use the graph to estimate the half-life of this substance.
Read the number of years when \(M = 0.5\).
The half-life is approximately 1.3 years.
Gradient as a rate of change
The gradient of a curve at any point is interpreted as the rate of change - a measure of how quickly one variable is changing as the other increases steadily.
Example
Using the graph from the previous question, find the rate of decay when \(x = 2\).
Solution
Draw a tangent to the curve at the point \(x = 2\).
\(\text{gradient of tangent} = \frac{0.72}{3.8} = 0.19\) (to two decimal places)
Rate of decay when \(x = 2\) is \(0.19\text{kg/year}\)
Decay is a negative change, therefore there should be a negative on the answer.
Test yourself
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