Families of Graphs
Theory
A family of graphs is a set of graphs whose equations have the same form, with one number, the parameter, changed. This Year 9 maths page shows how changing \(m\) or \(c\) changes a straight line, how \(a\) and \(c\) change a parabola, and how to use one point on a graph to find an unknown parameter. A graphing app with a slider is the quickest way to see each pattern. Curriculum: VC2M9A07 / AC9M9A06.
A parameter is a number in an equation that stays fixed for one graph but can be changed to make another. In \(y = mx + c\), \(x\) and \(y\) are the variables and \(m\) and \(c\) are parameters. Graphs drawn with the same equation but different parameter values make a family of graphs.
Straight lines \(y = mx + c\). The gradient \(m\) sets the steepness and direction: the bigger \(m\), the steeper the line, negative \(m\) slopes down from left to right and \(m = 0\) is horizontal. The \(y\)-intercept \(c\) is where the line crosses the \(y\)-axis. Lines with the same \(m\) are parallel. Lines with the same \(c\) all pass through \((0, c)\).
Parabolas \(y = ax^2 + c\). The turning point is \((0, c)\), so changing \(c\) moves the parabola up or down. If \(a > 0\) it is concave up; if \(a < 0\) it is concave down. The larger the size of \(a\) (ignoring the sign), the narrower the parabola.
Finding a parameter. If one point on a graph is known, substitute its coordinates into the equation to find an unknown parameter. For example, if \(y = kx^2 + 1\) passes through \((2, 9)\), then \(9 = 4k + 1\), so \(k = 2\).
Intercepts. Put \(x = 0\) to find the \(y\)-intercept and \(y = 0\) to find any \(x\)-intercepts. The parabola \(y = ax^2 + c\) meets the \(x\)-axis where \(ax^2 = -c\).
| Family | Change this | What happens |
|---|---|---|
| \(y = mx + c\) | \(m\) | steeper or flatter; negative slopes down |
| \(y = mx + c\) | \(c\) | parallel line moved up or down |
| \(y = ax^2 + c\) | \(a\) | narrower or wider; negative flips it |
| \(y = ax^2 + c\) | \(c\) | turning point \((0, c)\) moves |
Key points to find quickly:
So there are two \(x\)-intercepts when \(c < 0\), one when \(c = 0\) and none when \(c > 0\). For \(y = x^2 - 25\) they are \(x = \pm 5\).
For \(y = 3x - 12\) this is \(x = 4\). A horizontal line (\(m = 0\)) has no \(x\)-intercept unless it is the \(x\)-axis itself.
Substitute a known point to find \(a\) or \(c\). The parabola is symmetric about the \(y\)-axis, so \((-p, q)\) is on it too.
How to explore a family of graphs
- Name the parameter that changes and keep everything else fixed.
- Draw three or four members, by hand from a table of values or with a graphing app slider.
- Compare: what stays the same (a point, the shape, the gradient) and what changes (steepness, width, position, direction).
- Link each change to the number in the equation, and check it with one point by substituting.
- Generalise: write the pattern as a rule, such as “same \(m\) means parallel”.
How to match graphs to equations
- Use the shape: straight line or parabola.
- Use the sign: negative \(m\) slopes down and negative \(a\) is concave down.
- Use the intercept: put \(x = 0\) to find where each graph crosses the \(y\)-axis.
- Use steepness or width last, to separate graphs that still look alike.
How to find a parameter from a point
- Write the equation of the family with the unknown parameter in it.
- Substitute the \(x\)- and \(y\)-values of the known point.
- Solve for the parameter, then write the full equation of the graph.
- Check with another point, or use symmetry: \(y = ax^2 + c\) has the same \(y\)-value at \(x = p\) and \(x = -p\).
(i) Which point do all the lines pass through?
(ii) Write the line in the family that is parallel to \(y = 3x + 7\).
(iii) Find the value of \(m\) for the line that passes through \((2, 5)\).
(i) Every line has \(c = -1\), so every line crosses the \(y\)-axis at the same place:
| \(x\) | \(=\) | \(0\) |
| \(y\) | \(=\) | \(m \times 0 - 1 = -1\) |
All the lines pass through \(\boldsymbol{(0, -1)}\).
(ii) Parallel lines have the same gradient, so \(m = 3\). The line is \(\boldsymbol{y = 3x - 1}\).
(iii) Substitute \(x = 2\) and \(y = 5\):
| \(5\) | \(=\) | \(m \times 2 - 1\) |
| \(6\) | \(=\) | \(2m\) |
| \(\therefore\ m\) | \(=\) | \(\mathbf{3}\) |
Check: \(3 \times 2 - 1 = 5\). It is the same line as in part (ii).
(i) describe its shape and give the turning point.
(ii) find its \(x\)-intercepts.
(iii) Keep \(c = 12\). Find \(a\) so that \(y = ax^2 + 12\) has \(x\)-intercepts \(-4\) and \(4\). Is it wider or narrower than \(y = -3x^2 + 12\)?
(i) \(a = -3\) is negative, so it is concave down, and \(|-3| > 1\), so it is narrower than \(y = x^2\). With \(c = 12\), the turning point is \(\boldsymbol{(0, 12)}\), a maximum.
(ii) Put \(y = 0\):
| \(-3x^2 + 12\) | \(=\) | \(0\) |
| \(3x^2\) | \(=\) | \(12\) |
| \(x^2\) | \(=\) | \(4\) |
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{\pm 2}\) |
(iii) Put \(x = 4\), \(y = 0\):
| \(a \times 4^2 + 12\) | \(=\) | \(0\) |
| \(16a\) | \(=\) | \(-12\) |
| \(\therefore\ a\) | \(=\) | \(\boldsymbol{-\tfrac{3}{4}}\) |
The same turning point but intercepts further out, so \(y = -\tfrac{3}{4}x^2 + 12\) is wider. A smaller size of \(a\) means a wider parabola.
(i) Find \(k\).
(ii) Find the \(x\)-intercepts of the parabola.
(iii) Write down the other point on the parabola with \(y = 36\).
(i) Substitute \(x = 4\) and \(y = 36\):
| \(36\) | \(=\) | \(k \times 4^2 - 12\) |
| \(48\) | \(=\) | \(16k\) |
| \(\therefore\ k\) | \(=\) | \(\mathbf{3}\) |
The parabola is \(y = 3x^2 - 12\).
(ii) Put \(y = 0\):
| \(3x^2 - 12\) | \(=\) | \(0\) |
| \(x^2\) | \(=\) | \(4\) |
| \(\therefore\ x\) | \(=\) | \(\boldsymbol{\pm 2}\) |
(iii) The parabola is symmetric about the \(y\)-axis, so \(x = -4\) gives the same \(y\)-value. The point is \(\boldsymbol{(-4, 36)}\). Check: \(3 \times (-4)^2 - 12 = 48 - 12 = 36\).
(i) Write down the coordinates of \(B\).
(ii) Find \(A\) when \(m = -2\).
(iii) Find \(m\) so that triangle \(OAB\) has an area of \(27\) square units.
(i) Put \(x = 0\): \(y = 6\). Every line in the family passes through \(\boldsymbol{B(0, 6)}\).
(ii) Put \(y = 0\) in \(y = -2x + 6\):
| \(0\) | \(=\) | \(-2x + 6\) |
| \(2x\) | \(=\) | \(6\) |
| \(\therefore\ x\) | \(=\) | \(3\) |
So \(A\) is \(\boldsymbol{(3, 0)}\).
(iii) In general, \(0 = mx + 6\) gives \(x = -\dfrac{6}{m}\), which is positive because \(m < 0\). The triangle has base \(OA = -\dfrac{6}{m}\) and height \(OB = 6\):
| \(\tfrac{1}{2} \times \left(-\dfrac{6}{m}\right) \times 6\) | \(=\) | \(27\) |
| \(-\dfrac{18}{m}\) | \(=\) | \(27\) |
| \(\therefore\ m\) | \(=\) | \(\boldsymbol{-\tfrac{2}{3}}\) |
Check: \(m = -\tfrac{2}{3}\) gives \(x = -6 \div \left(-\tfrac{2}{3}\right) = 9\), and \(\tfrac{1}{2} \times 9 \times 6 = 27\).
Common pitfalls
Frequently asked questions
What is a family of graphs?
A set of graphs with the same form of equation where one number, the parameter, takes different values. For example, \(y = 2x + 1\), \(y = 2x + 3\) and \(y = 2x - 4\) are all in the family \(y = 2x + c\).
What does changing m do to y = mx + c?
It changes the steepness and direction of the line. A bigger \(m\) gives a steeper line, a negative \(m\) slopes down from left to right, and \(m = 0\) gives a horizontal line. The \(y\)-intercept stays at \(c\).
How can you tell if lines are parallel from their equations?
Write each in the form \(y = mx + c\). If the gradients \(m\) are the same, the lines are parallel. If the \(y\)-intercepts are also the same, they are the same line.
How do you find the x-intercepts of y = ax^2 + c?
Put \(y = 0\) and solve \(ax^2 = -c\). There are two \(x\)-intercepts when \(a\) and \(c\) have opposite signs, one (at the origin) when \(c = 0\), and none when \(a\) and \(c\) have the same sign. For example, \(y = 3x^2 - 300\) meets the \(x\)-axis at \(x = \pm 10\).
How do you find a parameter from a point on the graph?
Substitute the \(x\)- and \(y\)-values of the point into the equation, then solve for the parameter. For example, if \(y = mx + 7\) passes through \((2, 1)\), then \(1 = 2m + 7\), so \(m = -3\).