Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 08) Functions and Transformations cover? A: Interpret function notation, compose mappings, and relate algebraic changes to graph translations or stretches.
The core idea is simple: A function is an input-output rule; a transformation tells how its graph moves or stretches.
Use it as a working check: For composites, work from the inside function outward. For graph shifts, changes outside f(x) move the graph vertically, while changes inside the brackets move it horizontally.
Then go one layer deeper: Use the examples to track each step clearly: substitute inputs, compose functions in the stated order, reverse operations for an inverse, and describe graph movement in plain words.
Function language appears in calculator prompts, inverse-relationship questions, and graph sketching. Track how algebraic edits move or stretch the curve.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
Before describing a graph transformation, locate where the change is made.
Algebraic change
Where it acts
Graph effect
Trap to avoid
f(x)+k
Outside the function output
Move every point up by k.
Do not change the input values.
f(x)−k
Outside the function output
Move every point down by k.
Do not call this a horizontal shift.
f(x−a)
Inside the input bracket
Move the graph right by a.
The sign feels reversed because the input must be larger to give the same bracket value.
f(x+a)
Inside the input bracket
Move the graph left by a.
Do not read the plus sign as "right".
af(x)
Outside the function output
Stretch vertical distances by factor a.
This changes y-values, not x-values.
f(ax)
Inside the input bracket
Compress horizontal distances by factor a, for a>1.
This changes x-values, not y-values.
Misconception check: outside changes affect outputs, so they move or stretch vertically. Inside changes affect inputs, so they move or stretch horizontally.
Inside multiplier checkpoint
For changes inside the bracket, keep the output value fixed and ask which new input gives the same old input.
New graph
Input condition for the same output
Point movement
Common trap
y=f(2x)
Need 2x=4 to use the old input 4.
A point (4,7) moves to (2,7).
Multiplying the x-coordinate by 2 instead of dividing it by 2.
y=f(x/2)
Need x/2=4 to use the old input 4.
A point (4,7)
y=f(−x)
Need −x=4 to use the old input 4.
A point (4,7)
Worked check: if (4,7) lies on y=f(x), then f(4)=7. On y=f(2x), set 2x=4, so x=2 and the matching point is (2,7). On y=f(x/2), set x/2=4, so x=8 and the matching point is (8,7).
Misconception check: inside multipliers change x-values in the reciprocal direction. The y-value stays 7 in the worked check because the output of f has not been multiplied.
Composite function order checkpoint
For composite notation, read the function closest to x first. Then feed that output into the outside function.
Expression
First step
Second step
Common trap
(f∘g)(x)
Find g(x).
Substitute the result into f.
Doing f(x) first because f is written on the left.
(g∘f)(x)
Find f(x).
Substitute the result into g.
Assuming (f∘g)(x)
f(g(2))
Evaluate g(2).
Put that number into f.
Substituting 2 into both functions separately.
Worked check: if f(x)=2x−3 and g(x)=x2+1, then
(f∘g)(x)=f(g(x))=2(x2+1)−3=2x2−1.
Misconception check: the composition symbol does not mean multiplication. (f∘g)(x) means "do g, then do f", not f(x)×g(x).
Worked example - Transformation tracker
The graph of y=x2 is transformed to y=3(x−2)2+5. Describe the sequence of transformations.
Start with y=x2.
x↦x−2 translates the graph 2 units to the right.
Multiplying by 3 stretches it vertically by factor 3 (parabola becomes narrower).
Adding 5 shifts the graph up by 5 units.
So the sequence is: translate right by 2, stretch vertically by 3, then shift up by 5.
Worked example - Compose functions carefully
Let f(x)=2x−3 and g(x)=x2+1. Find (f∘g)(x) and evaluate (g∘f)(2).
Compute (f∘g)(x)=f(g(x)):
First find g(x): x2+1.
Substitute into f: f(g(x))=2(x2+1)−3=2x2+2−3=2x2−1
Compute (g∘f)(2)=g(f(2)):
f(2)=2(2)−3=1
Thus (f∘g)(x)=2x2−1 and (g∘f)(2)=2.
Worked example - Find an inverse function
Given h(x)=23x−5, determine h−1(x) and state the transformation that links h and h−1.
Let y=23x−5.
Swap x and y: x=23y−5. This reflects that an inverse reverses the original mapping-the output y from h becomes the input for h−1, while the original input x becomes the output we now solve for.
Solve for y:
Multiply by 2: 2x=3y−5.
Add 5: 3y=2x+5.
Divide by 3: y=32x+5
Therefore h−1(x)=32x+5.
The original function multiplies by 3, subtracts 5, then divides by 2. The inverse reverses these steps: multiply by 2, add 5, then divide by 3. Graphically, h and h−1 are reflections of each other across the line y=x.
Inverse function check checkpoint
After finding an inverse, check that it reverses the original mapping. This catches sign errors that still look algebraically neat.
Check
What to do
What it proves
Trap to avoid
Operation check
List the operations in f, then reverse them in the opposite order for f−1.
The inverse undoes each step.
Reversing the operations but keeping them in the original order.
Composition check
Compute f(f−1(x)) or f−1(f(x)).
A correct inverse gives back x.
Stopping after solving for y without checking the result.
Graph check
Compare the graphs of y=f(x) and y=f−1(x).
They should reflect in the line y=x
Worked check: for h(x)=23x−5 and h−1(x)=32x+5, substitute back:
h(h−1(x))=23(32x+5)−5=22x+5−5=x.
Misconception check: an inverse function is not the same as a reciprocal. f−1(x) reverses the mapping; it does not mean f(x)1.
Practice Quiz
Consolidate function evaluation, compositions, and transformation language with auto-marked prompts.
Try this
Given f(x)=2x−1 and g(x)=x2+3, find (g∘f)(x) and state the value when x=−2.
