Before reading this guide, it may be helpful to read the guide from M3 on coordinates and graphs.
Quadratic graphs
An equation of the form \(y = mx + c\) where \(m\) and \(c\) are numbers, gives a straight line when values of \(x\) and corresponding values of \(y\) are plotted on a grid.
A straight line is a linear graph.
A quadratic graph has an equation which includes an \(x^{2}\) term.
Examples of equations that form quadratic graphs:
\(y = x^{2}\)
\(y = 3x^{2} + 2x – 5\)
\(y = 9 – x^{2}\)
\(y = 4x – 6x^{2}\)
To draw a quadratic graph from its equation,
- Set up or complete a table with values of \(x\)
- Substitute each value of \(x\) into the equation to find each corresponding value of \(y\)
- Plot the \(x\) and \(y\) values as coordinates
- Join the points with a smooth curve
The table is usually already set up but may have some missing values to be filled in. The grid will also be set up with an appropriate scale on both axes. Typically, you have to plot the points from the table and join them with a smooth curve.
Completing quadratic graphs
Example
Complete the table for the equation \(y = x^{2} – 4x + 3\) and draw its graph for values of \(x\) from –2 to 5.
| \(x\) | –2 | –1 | 0 | 1 | 2 | 3 | 4 | 5 |
| \(y\) | 15 | 8 | 0 | –1 | 0 | 8 |
Solution
Substitute \(x =0\) and \(x =4\) into the equation to find values of \(y\).
When \(x =0\), \(y =3\) (\(x^{2}\) and \(–4x\) are both \(0)\)
When \(x =4\), \(y =4^{2} – 4(4) = 16 – 16 + 3 = 3\)
| \(x\) | –2 | –1 | 0 | 1 | 2 | 3 | 4 | 5 |
| \(y\) | 15 | 8 | 3 | 0 | –1 | 0 | 3 | 8 |
Plot the \(x\) and \(y\) values as coordinates and join them with a smooth curve.
This is a typical shape for a quadratic function. Always join points freehand with a curve - joining points with a ruler will lose marks! Your curve can be a bit ragged but will still get the marks.
Question
Complete the table for the equation \(y = 2x^{2} +2x – 4\) and draw its graph for values of \(x\) from –3 to 2.
| \(x\) | –3 | –2 | –1 | 0 | 1 | 2 |
| \(y\) | 8 | –4 | –4 | 0 | 8 |
When \(x = –2\), \(y = 2(–2)^{2} + 2(–2) –4\)
- When evaluating \(2x^{2}\), calculate \(x^{2}\) first and then multiply by \(2\).
\( = 8 – 4 + 4 \)
\( = 0 \)
When \(x = 2\), \(y = 2(2)^{2} + 2(2) –4\)
\( = 8 + 4 – 4 = 8 \)
| \(x\) | –3 | –2 | –1 | 0 | 1 | 2 |
| \(y\) | 8 | 0 | –4 | –4 | 0 | 8 |
Solving equations with quadratic graphs
Graphs of quadratic functions can be used to solve equations.
Example
Draw the graph of \(y = 2x^{2} + 2x – 4\) for values of –3 to 2. Use the graph to solve the equation \(y = 2x^{2} + 2x – 4\).
Solution
Let's use the graph from the previous question, which illustrates the equation \(y = 2x^{2} + 2x – 4\).
The graph is of the function \(y = 2x^{2} + 2x – 4\).
We want to solve the equation \( 2x^{2} + 2x – 4 = 0\).
In the equation \( 2x^{2} + 2x – \ 4 = 0\), we know that \( 2x^{2} + 2x – \ 4\) is equal to \(y\).
We need to draw the line \(y = 0\) which is the \(x\)-axis.
The points of intersection of the \(x\)-axis and the curve give the solutions to \(2x^{2} + 2x – 4 = 0\).
The axis and curve intersect when \(x = –2\) and \(x = 1\).
Solutions to the equation \(2x^{2} + 2x – 4 = 0\) are \(x = –2\) and \(x = 1\).
Note that there are two solutions since the curve and \(x\)-axis intersect at two points.
Example
Using the graph from the previous example to solve the equation \(2x^{2} + 2x – 4 = -4\).
Hint – the equation \(2x^{2} + 2x – 4 = y\) is the same as \(2x^{2} + 2x – 4 = -4\). Therefore \(y = -4\).
Solution
First, we need to draw the line \(y = –4\).
The points of intersection of the line \(y = –4\) and the curve give the solutions to \(2x^{2} + 2x – 4 = –4\).
The line and curve intersect when \(x = –1\) and \(x = 0\).
Solutions to the equation \(2x^{2} + 2x – 4 = –4\) are \(x = –1\) and \(x = 0\).
Question
Complete the table for the equation \(y = x^{2} – x – 2\) and draw its graph for values of \(x\) from –4 to 4.
| \(x\) | –4 | –3 | –2 | –1 | 0 | 1 | 2 | 3 | 4 |
| \(y\) | 18 | 10 | 0 | –2 | 0 | 4 |
Answer
First, you will need a grid like this to plot a graph.
- When \(x = –2\), \(y = (–2)^{2} – (–2) – 2\)
\(y = 4 + 2 – 2\)
\(= 4\)
- When \(x = 0\), \(y = 0 – 0 – 2\)
\(= –2\)
- When \(x = 4\), \(y = (4)^{2} – (4) – 2\)
\(y = 16 – 4 – 2\)
\(= 10\)
| \(x\) | –4 | –3 | –2 | –1 | 0 | 1 | 2 | 3 | 4 |
| \(y\) | 18 | 10 | 4 | 0 | –2 | –2 | 0 | 4 | 10 |
With these, we can now plot the points and draw a line through them, which should appear curved.
Question
Using the graph from the question above, draw the line \(y = 6\) on the grid.
Write down the \(x\) values of the points of the intersection of this line and the curve.
Read from the dashed lines – this will be an approximate answer as the readings are between integers.
Where the dashed lines meet the \(x\)-axis gives approximately \(x = –2.3\) and \(x = 3.3\).
More on M6: Algebra
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