Determining Rules Revision Notes for VCE SSCE Mathematical Methods

Determining Rules

Introduction to determining rules

When working with different types of graphs, we need specific information to write their equations. This information is called sufficient conditions – it means we have just enough detail to determine the rule completely.

Previously, you learned about sufficient conditions for straight lines and parabolas. For straight lines, you could determine the equation if you knew:

The coordinates of two points on the line
The gradient and one point on the line

For parabolas, you could determine the equation if you knew:

The coordinates of three points on the parabola
The coordinates of the vertex and one other point

In this section, we'll explore sufficient conditions for determining the rules of other graph types, including rectangular hyperbolas, square root functions, and circles.

General approach to determining rules

The key technique for finding unknown parameters in an equation is substitution. Here's the general process:

Identify the general form of the equation for your graph type
Determine which parameters are unknown
Substitute the coordinates of known points into the equation
If you have one unknown, solve the resulting equation
If you have multiple unknowns, create simultaneous equations and solve them together

infoNote

The number of unknowns determines how many points you need: one unknown requires one point, while two unknowns require two points to create simultaneous equations.

Let's look at specific examples for different graph types.

Rectangular hyperbolas

The general form of a rectangular hyperbola is:

y = \frac{a}{x-h} + k

where $a$ , $h$ , and $k$ are parameters that determine the shape and position of the hyperbola.

Finding one unknown parameter

When you have only one unknown parameter and one point, direct substitution gives you a simple equation to solve.

lightbulbExample

Worked Example: Finding One Unknown Parameter

Question: The rectangular hyperbola $y = \frac{a}{x} + 8$ passes through the point $(-2, 6)$ . Find the value of $a$ .

Solution:

Since the point $(-2, 6)$ lies on the hyperbola, we can substitute $x = -2$ and $y = 6$ into the equation.

6 = \frac{a}{-2} + 8

6 - 8 = \frac{a}{-2}

-2 = \frac{a}{-2}

a = 4

Therefore, the equation is $y = \frac{4}{x} + 8$ .

Finding two unknown parameters

When dealing with two unknowns, you'll need to use simultaneous equations. This requires substituting two different points to create two separate equations.

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Worked Example: Finding Two Unknown Parameters

Question: The rectangular hyperbola $y = \frac{a}{x} + k$ passes through the points $(2, 7)$ and $(-1, 1)$ . Find the values of $a$ and $k$ .

Solution:

Since we have two unknowns ( $a$ and $k$ ), we need two equations. We create these by substituting both points.

When $x = 2$ , $y = 7$ :

7 = \frac{a}{2} + k \quad \text{...(1)}

When $x = -1$ , $y = 1$ :

1 = \frac{a}{-1} + k

1 = -a + k \quad \text{...(2)}

Now we solve these simultaneous equations. Subtract equation (2) from equation (1):

7 - 1 = \frac{a}{2} + k - (-a + k)

6 = \frac{a}{2} + a

Multiply both sides by 2:

12 = a + 2a

12 = 3a

a = 4

Substitute $a = 4$ into equation (2):

1 = -4 + k

k = 5

Therefore, the equation is $y = \frac{4}{x} + 5$ .

chatImportant

When solving simultaneous equations, look for opportunities to eliminate one variable through addition or subtraction. This simplifies the algebra significantly.

Square root functions

The general form of a square root function is:

y = a\sqrt{x-h} + k

where $a$ , $h$ , and $k$ are parameters.

Square root functions require careful algebraic manipulation because you may need to square both sides of an equation. This technique can introduce extraneous solutions, so always check your work.

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Worked Example: Square Root Function with Two Unknowns

Question: A graph with rule $y = a\sqrt{x-h}$ passes through the points $(4, 2)$ and $(7, 4)$ . Find the values of $a$ and $h$ .

Solution:

Substitute both points to create two equations.

When $x = 4$ , $y = 2$ :

2 = a\sqrt{4-h} \quad \text{...(1)}

When $x = 7$ , $y = 4$ :

4 = a\sqrt{7-h} \quad \text{...(2)}

Divide equation (2) by equation (1) to eliminate $a$ :

\frac{4}{2} = \frac{a\sqrt{7-h}}{a\sqrt{4-h}}

2 = \frac{\sqrt{7-h}}{\sqrt{4-h}} \quad \text{...(3)}

Multiply both sides by $\sqrt{4-h}$ :

2\sqrt{4-h} = \sqrt{7-h}

Square both sides:

4(4-h) = 7-h

16 - 4h = 7 - h

16 - 7 = 4h - h

9 = 3h

h = 3

Substitute $h = 3$ into equation (1):

2 = a\sqrt{4-3}

2 = a\sqrt{1}

2 = a

Therefore, the equation is $y = 2\sqrt{x-3}$ .

infoNote

For square root functions, dividing one equation by another can simplify the algebra considerably. Remember that when you square both sides, you must ensure your solution doesn't create any undefined expressions in the original equations.

Circles

The centre-radius form for the equation of a circle is:

(x-h)^2 + (y-k)^2 = r^2

where $(h, k)$ is the centre of the circle and $r$ is the radius.

When finding the equation of a circle, remember that any point on the circle is exactly one radius length from the centre. This fundamental property allows us to use the distance formula to find the radius.

lightbulbExample

Worked Example: Circle Given Centre and a Point

Question: Find the equation of the circle whose centre is at the point $(1, -1)$ and which passes through the point $(4, 3)$ .

Solution:

We know the centre is $(1, -1)$ , so $h = 1$ and $k = -1$ . We need to find the radius $r$ .

Since the point $(4, 3)$ lies on the circle, the distance from the centre to this point equals the radius. Use the distance formula:

r = \sqrt{(4-1)^2 + (3-(-1))^2}

r = \sqrt{3^2 + 4^2}

r = \sqrt{9 + 16}

r = \sqrt{25}

r = 5

Substitute into the centre-radius form:

(x-1)^2 + (y-(-1))^2 = 5^2

(x-1)^2 + (y+1)^2 = 25

chatImportant

For circles, if you know the centre and one point on the circle, use the distance formula to find the radius. The distance from the centre to any point on the circle is always equal to the radius.

Key takeaways

bookmarkSummary

Remember These Key Points:

Substitution is the foundation: Always substitute known coordinates into the general form of the equation.
Match unknowns with equations: If you have one unknown parameter, you need one point. If you have two unknowns, you need two points to create simultaneous equations.
Different graphs require different techniques:
- Rectangular hyperbolas often need straightforward algebraic manipulation
- Square root functions may require dividing equations and squaring both sides
- Circles use the distance formula to find the radius
For circles, use the centre-radius form: Remember that any point on the circle is exactly one radius length from the centre. Start with $(x-h)^2 + (y-k)^2 = r^2$ .
Always verify your answer: After finding parameters, check that your equation works by substituting the original points back in.

Determining Rules (VCE SSCE Mathematical Methods): Revision Notes