Determining Rules
Learn to determine the rule of a graph for Queensland Year 11 Mathematical Methods (QCAA): working backwards from a curve to find its equation, whether a parabola, hyperbola, square-root graph or circle.
You will learn to name the family, read the turning point, asymptotes, endpoint or centre, then use one further point to pin down the parameters — a supporting skill that links each graph back to its equation.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), determining the rule means working backwards from a graph to its equation. This page shows how to read the key features of a parabola, hyperbola, square-root graph or circle, and use a further point to pin down the constants.
Determining the rule is the reverse of sketching: you are given a graph or its features and must find the equation. The trick is to start from the form of the family, read the constants you can see, then use one more point to find what is left.
Each family shows its key features directly: a parabola \(y=a(x-h)^2+k\) reveals its turning point \((h,\,k)\); a hyperbola \(y=\dfrac{a}{x-h}+k\) reveals its asymptotes \(x=h,\ y=k\); a square-root graph \(y=a\sqrt{x-h}+k\) reveals its endpoint \((h,\,k)\); a circle \((x-h)^2+(y-k)^2=r^2\) reveals its centre and radius.
The standard forms whose features you read from the graph:
The circle, with centre the midpoint of a diameter:
How to determine a rule from a graph
- Identify the family: parabola, hyperbola, square-root or circle, from the overall shape.
- Read the fixed features: turning point, asymptotes, endpoint or centre give \(h\) and \(k\).
- Use a point: substitute a further point on the curve and solve for the remaining constant \(a\) (or the radius \(r\)).
Frame — use the turning-point form:
| \(y\) | \(=\) | \(a(x-2)^2-1\) |
Scale — substitute \((0,\,3)\):
| \(3\) | \(=\) | \(a(0-2)^2-1\) |
| \(3\) | \(=\) | \(4a-1\) |
| \(4a\) | \(=\) | \(4\) |
| \(a\) | \(=\) | \(1\) |
Rule: \(y=(x-2)^2-1\).
Frame — the asymptotes give \(h\) and \(k\):
| \(y\) | \(=\) | \(\dfrac{a}{x-1}-2\) |
Scale — substitute \((2,\,1)\):
| \(1\) | \(=\) | \(\dfrac{a}{2-1}-2\) |
| \(1\) | \(=\) | \(a-2\) |
| \(a\) | \(=\) | \(3\) |
Rule: \(y=\dfrac{3}{x-1}-2\).
Frame — the endpoint gives \(h\) and \(k\):
| \(y\) | \(=\) | \(a\sqrt{x-1}+2\) |
Scale — substitute \((5,\,4)\):
| \(4\) | \(=\) | \(a\sqrt{5-1}+2\) |
| \(4\) | \(=\) | \(a\sqrt{4}+2\) |
| \(4\) | \(=\) | \(2a+2\) |
| \(2a\) | \(=\) | \(2\) |
| \(a\) | \(=\) | \(1\) |
Rule: \(y=\sqrt{x-1}+2\).
Centre — the midpoint of the diameter:
| \((h,\,k)\) | \(=\) | \(\left(\dfrac{1+5}{2},\,\dfrac{2+8}{2}\right)\) |
| \(=\) | \((3,\,5)\) |
Radius — from the centre to an endpoint:
| \(r^2\) | \(=\) | \((5-3)^2+(8-5)^2\) |
| \(=\) | \(2^2+3^2\) | |
| \(=\) | \(4+9\) | |
| \(=\) | \(13\) |
Equation:
| \((x-3)^2+(y-5)^2\) | \(=\) | \(13\) |
Equation: \((x-3)^2+(y-5)^2=13\).
Common pitfalls
Frequently asked questions
How do you find the rule of a graph?
Identify the family and its standard form, read the fixed features (turning point, asymptotes, endpoint or centre) to get \(h\) and \(k\), then substitute a further point to solve for the remaining constant.
How do you find a in y equals a times x minus h squared plus k?
Read the turning point \((h,\,k)\) from the graph, substitute a second known point, and solve the resulting linear equation for \(a\).
How do you determine the rule of a hyperbola from a graph?
The asymptotes give \(x=h\) and \(y=k\), so the rule is \(y=\dfrac{a}{x-h}+k\); substitute a point on the curve to find \(a\).
How do you find the equation of a circle from a diameter?
The centre is the midpoint of the diameter and the radius is half its length, so compute both and substitute into \((x-h)^2+(y-k)^2=r^2\).
Why do you need an extra point when determining a rule?
The visible features fix the position of the curve, but not its steepness; the extra point determines the scale factor \(a\) (or the radius).