Welcome to Formulae and Functions
Welcome to your study guide for Formulae and Functions in CCEA GCSE Mathematics (2210). Whether you are sitting the Foundation Tier (\(M1, M2, M5, M6\)) or the Higher Tier (\(M3, M4, M7, M8\)), this topic gives you the tools to express real-world patterns in algebra, predict values, and rearrange rules to find missing information.
Don't worry if algebra has felt confusing in the past! We will break every concept down into simple, step-by-step recipes so you can tackle exam questions with total confidence.
1. Key Definitions: What is the Difference?
In GCSE Maths, words like expression, equation, and formula have very specific meanings. Let's make sure you never mix them up:
1. Expression: A collection of terms with no equals sign or inequality sign.
Example: \(3x + 5\) or \(4a - 2b\)
2. Equation: An algebraic statement equating two expressions containing an unknown letter to solve for a specific value.
Example: \(2x + 3 = 11\) (Here, \(x\) has one specific value: \(x = 4\)).
3. Formula (plural: Formulae): A mathematical rule showing the relationship between two or more distinct named physical or abstract variables using an equals sign.
Example: \(v = u + at\) (linking velocity, initial speed, acceleration, and time) or \(A = \pi r^2\) (area of a circle).
4. Function: A mathematical relationship where every permitted input value (\(x\)) produces a single, uniquely determined output value (\(y\) or \(f(x)\)).
5. Subject of a Formula: The single variable that stands completely alone on one side of the equals sign with a positive coefficient of \(+1\) and a power of \(1\).
Example: In \(y = mx + c\), the subject is \(y\).
Key Takeaway: A formula links two or more real-life variables (like speed and time). An equation is solved to find a specific mystery number. An expression has no equals sign at all.
2. Function Machines: Working Forwards and Backwards
Think of a function machine like a kitchen blender: you put ingredients in (Input), the machine performs a set of actions in order (Operations), and out comes a smoothie (Output).
Working Forwards (Finding the Output): Follow the operations from left to right in the exact order shown.
Example: Input \(\rightarrow\) Add 3 \(\rightarrow\) Multiply by 4 \(\rightarrow\) Output.
If the input is \(x = 5\):
Step 1: \(5 + 3 = 8\)
Step 2: \(8 \times 4 = 32\)
The output is \(32\). In algebraic form, this function is \(\text{Output} = 4(x + 3)\).
Working Backwards (Finding the Input): If you know the output and need to find the starting input, you must run the machine in reverse using inverse (opposite) operations.
Example: If the output is \(40\) for the machine above:
Step 1: Reverse the last operation (Divide by \(4\)): \(40 \div 4 = 10\)
Step 2: Reverse the first operation (Subtract \(3\)): \(10 - 3 = 7\)
The original input was \(7\). Check: \((7 + 3) \times 4 = 10 \times 4 = 40\).
Quick Review of Inverse Pairs:
Addition (\(+\)) \(\leftrightarrow\) Subtraction (\(-\))
Multiplication (\(\times\)) \(\leftrightarrow\) Division (\(\div\))
Squaring (\(x^2\)) \(\leftrightarrow\) Square rooting (\(\sqrt{x}\))
3. Families of Functions and Their Graphs
In CCEA GCSE Maths, you need to recognise the shape and algebraic form of five core function families:
1. Linear Functions: \(y = mx + c\)
These produce a straight-line graph. \(m\) is the gradient (steepness), calculated as \(m = \frac{y_2 - y_1}{x_2 - x_1}\), and \(c\) is the \(y\)-intercept (where the line crosses the vertical \(y\)-axis).
2. Quadratic Functions: \(y = ax^2 + bx + c\) (where \(a \neq 0\))
These produce a smooth U-shaped or n-shaped curve called a parabola. It has a single turning point (vertex) and a vertical line of symmetry.
3. Cubic Functions: \(y = ax^3 + bx^2 + cx + d\) (where \(a \neq 0\))
These produce characteristic S-shaped curves that can have up to two turning points.
4. Reciprocal Functions: \(y = \frac{k}{x}\) (where \(x \neq 0\))
These produce a curve called a hyperbola that gets closer and closer to the \(x\)-axis and \(y\)-axis without ever touching them (the axes are asymptotes).
5. Exponential Functions: \(y = k^x\) or \(y = A \times b^x\) (where \(b > 0, b \neq 1\))
These model rapid growth or decay (such as population growth or radioactive decay), curving steeply upwards or levelling off towards zero.
4. Deriving Formulae and Substitution
Deriving a Formula from Words:
Read the real-world scenario and turn the words into letters.
Example: A plumber charges a fixed call-out fee of \(F\) pounds plus an hourly rate of \(r\) pounds for \(t\) hours worked. Write a formula for the total cost \(C\).
Answer: \(C = F + rt\)
Substitution (Plugging in Numbers):
Substitution means replacing algebraic letters with given numbers. Always follow the priority of operations: BIDMAS (Brackets, Indices/Powers, Division/Multiplication, Addition/Subtraction).
Golden Rule for Negative Numbers: Always put negative numbers in brackets when substituting, especially when squaring!
Worked Example 1: Find the value of \(P = 3x^2 - 5x\) when \(x = -4\).
Step 1: Write it with brackets: \(P = 3(-4)^2 - 5(-4)\)
Step 2: Work out powers first: \((-4)^2 = 16\)
Step 3: Multiply: \(3 \times 16 = 48\), and \(-5 \times -4 = +20\)
Step 4: Add: \(P = 48 + 20 = 68\)
Common Examiner Trap: Entering \(-4^2\) into a calculator without brackets gives \(-16\). But \((-4)^2 = (-4) \times (-4) = +16\)!
5. Changing the Subject of a Formula (Rearranging)
Rearranging a formula means shuffling the letters so that a different variable is left entirely alone on one side of the equals sign. Whatever you do to one side of the equation, you must do to the other side!
Level 1: Linear One-Step and Two-Step Formulae
Example: Make \(x\) the subject of \(y = mx + c\).
Step 1: Subtract \(c\) from both sides to isolate the \(x\)-term: \(y - c = mx\)
Step 2: Divide both sides by \(m\): \(x = \frac{y - c}{m}\)
Level 2: Formulae Involving Powers and Roots
Always isolate the power or square root expression first before squaring or square rooting the entire side.
Example A (Powers): Make \(r\) the subject of the surface area formula \(A = 4\pi r^2\).
Step 1: Divide both sides by \(4\pi\): \(r^2 = \frac{A}{4\pi}\)
Step 2: Take the square root of both sides: \(r = \sqrt{\frac{A}{4\pi}}\)
Example B (Roots): Make \(t\) the subject of \(v = \sqrt{\frac{2s}{t}}\).
Step 1: Square both sides to eliminate the root: \(v^2 = \frac{2s}{t}\)
Step 2: Multiply both sides by \(t\) to get it out of the denominator: \(t \cdot v^2 = 2s\)
Step 3: Divide both sides by \(v^2\): \(t = \frac{2s}{v^2}\)
Level 3: Subject Appearing in More Than One Term (Higher Tier)
When the letter you want appears in multiple places (often in a fraction), follow this reliable 5-Step Method:
Step 1: Clear fractions by multiplying every term by the denominator.
Step 2: Expand all brackets to release the target variable.
Step 3: Collect all terms containing the new subject on one side, and move all other terms to the opposite side.
Step 4: Factorise the target variable out as a common factor.
Step 5: Divide by the remaining bracket.
Worked Example: Make \(x\) the subject of \(y = \frac{ax + b}{cx + d}\).
Step 1 (Clear fraction): Multiply by \((cx + d)\) \(\implies y(cx + d) = ax + b\)
Step 2 (Expand): \(cxy + dy = ax + b\)
Step 3 (Collect \(x\) terms on the left, others on the right): \(cxy - ax = b - dy\)
Step 4 (Factorise out \(x\)): \(x(cy - a) = b - dy\)
Step 5 (Divide by bracket): \(x = \frac{b - dy}{cy - a}\)
Key Takeaway: If the target letter appears twice, you will almost always need to factorise it out in the second-to-last step!
6. Direct and Inverse Proportionality Formulae (Higher Tier)
Proportionality problems ask you to find the mathematical rule connecting two quantities using a constant of proportionality, \(k\).
Direct Proportion (\(y \propto x^n\))
As \(x\) increases, \(y\) increases at a corresponding rate.
Setup: Replace the proportionality symbol \(\propto\) with \(= k\), so \(y = kx^n\).
Worked Example: \(y\) is directly proportional to the square of \(x\). When \(x = 3\), \(y = 36\). Find \(y\) when \(x = 5\).
Step 1 (Write the equation): \(y = kx^2\)
Step 2 (Substitute known values to find \(k\)): \(36 = k \times 3^2 \implies 36 = 9k \implies k = 4\)
Step 3 (Write the specific formula): \(y = 4x^2\)
Step 4 (Calculate the new value): When \(x = 5\), \(y = 4 \times 5^2 = 4 \times 25 = 100\).
Inverse Proportion (\(y \propto \frac{1}{x^n}\))
As \(x\) increases, \(y\) decreases.
Setup: Replace \(\propto\) with \(= \frac{k}{\dots}\), so \(y = \frac{k}{x^n}\).
Worked Example: The intensity \(I\) is inversely proportional to the square of distance \(d\). When \(d = 2\), \(I = 15\). Find \(I\) when \(d = 5\).
Step 1 (Write the equation): \(I = \frac{k}{d^2}\)
Step 2 (Find \(k\)): \(15 = \frac{k}{2^2} \implies 15 = \frac{k}{4} \implies k = 15 \times 4 = 60\)
Step 3 (Write the specific formula): \(I = \frac{60}{d^2}\)
Step 4 (Calculate the new value): When \(d = 5\), \(I = \frac{60}{5^2} = \frac{60}{25} = 2.4\).
7. Pitfalls to Avoid in the Exam
Examiners frequently report these common mistakes in CCEA papers. Watch out for them!
1. The Squaring Trap in Substitution:
For \(3x^2\) when \(x = -2\), do NOT calculate \((3 \times -2)^2 = (-6)^2 = 36\). BIDMAS says evaluate the power first: \(3 \times (-2)^2 = 3 \times 4 = 12\).
2. Incomplete Division Across Terms:
If \(ay = bx + c\), isolating \(y\) requires dividing the entire side: \(y = \frac{bx + c}{a}\), NOT \(y = \frac{bx}{a} + c\).
3. Leaving the Subject on Both Sides:
An answer like \(x = \frac{2x + 5}{y}\) gets zero marks for rearranging. The subject \(x\) must only appear once in the entire final line.
4. Partial Square Rooting:
If \(a^2 = b^2 + c^2\), you cannot square root term-by-term to say \(a = b + c\). The correct rearrangement is \(a = \sqrt{b^2 + c^2}\).
5. Upside-Down Fraction Error:
If \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\), you cannot simply flip individual terms to say \(R = R_1 + R_2\). You must first combine the right-hand side over a common denominator before inverting.
Quick Summary Checklist
Before moving on, make sure you can:
1. Distinguish clearly between an expression, an equation, and a formula.
2. Work backwards through a function machine using inverse operations.
3. Identify linear, quadratic, cubic, reciprocal, and exponential graphs.
4. Substitute positive and negative numbers accurately using BIDMAS.
5. Rearrange linear formulae, power/root formulae, and fractions where the subject appears twice.
6. Set up and solve direct and inverse proportionality equations using the constant \(k\).