Graph Transformations (AA SL)
Once you can read \(y=f(x)\), the next skill is predicting what happens when it's translated, reflected or stretched - without ever needing to know what \(f\) actually is. This page covers the four single transformations and how to chain them correctly, as part of the wider Graphs & Transformations topic.
30 questions on this sub-topic.
Building composite transformations
Covered under IB syllabus reference SL2.11. These are prior-knowledge relationships rather than formula-booklet entries - the booklet won't state them, so you need to know them cold.
Composite transformations
\[y=x^2 \;\longrightarrow\; y=3x^2+2\]
Apply transformations in the order the function is built: stretch by 3, then translate 2 up.
Not in the formula booklet - prior knowledgeThe four single moves
\(y=f(x)+b,\ y=f(x-a),\ y=-f(x),\ y=pf(x)\)
Translate up/down, translate left/right, reflect in the \(x\)-axis, and stretch vertically by scale factor \(p\).
Need the full syllabus wording or how this connects to specific function types? See Graphs & Transformations.
Worked examples
Describe the single transformation mapping \(y=f(x)\) to:
(a) \(y=f(x)-5\)
(b) \(y=f(x+3)\)
Worked solution
(a) \(f(x)-5\): subtracting from the output is a translation \(5\) units down. A1
Translation \(5\) units down. A1
(b) \(f(x+3)\): replacing \(x\) by \(x+3\) is a translation \(3\) units to the left. A1
Translation \(3\) units to the left. A1
The graph of \(y=f(x)\) is transformed to \(y=3f(x)-2\). The point \((4,-1)\) is on \(y=f(x)\).
(a) Find the image of \((4,-1)\).
(b) If \(f\) has a minimum value of \(-1\), find the minimum value of \(3f(x)-2\).
Worked solution
(a) \(y=3f(x)-2\) leaves \(x\) unchanged and maps the \(y\)-value: \(y=3(-1)-2\). M1
\(=-5\), so \((4,-1)\to(4,-5)\). A1
(b) A minimum maps the same way (the stretch factor 3 is positive, so order is preserved): minimum \(=3(-1)-2\). M1
Since a positive stretch preserves the minimum, this equals \(-5\). R1
Minimum value \(=-5\). A1
Let \(f(x)=\sqrt{x}\) with domain \(x\ge0\) and range \(f(x)\ge0\). For \(g(x)=\sqrt{x-1}+2\):
(a) State the domain.
(b) State the range.
Worked solution
(a) The radicand of \(\sqrt{x-1}\) must be \(\ge0\): \(x-1\ge0\Rightarrow x\ge1.\) M1
\(x\ge1.\) A1
(b) \(\sqrt{x-1}\ge0\). R1
The \(+2\) shifts the output up 2 units. A1
\(g(x)\ge2.\) A1
Common mistakes
- Applying transformations in the wrong order. \(y=3f(x)+2\) means "stretch by 3, then translate up 2" - doing the translation first gives a different (wrong) graph.
- Getting the direction of a horizontal translation backwards. \(y=f(x-a)\) shifts the graph right by \(a\) when \(a>0\), not left - the sign inside the bracket is the opposite of the direction of movement.
- Confusing \(y=pf(x)\) with \(y=f(px)\). The first stretches vertically by factor \(p\); the second stretches horizontally by factor \(\tfrac1p\) - multiplying \(x\) inside the brackets compresses the graph when \(p>1\), it doesn't stretch it.
Ready to practise properly?
30 graph-transformation questions, marked instantly like the real exam.
Quick answers
How do I know which way a horizontal translation moves the graph?
\(y=f(x-a)\) shifts the graph right by \(a\) when \(a\) is positive, and left when \(a\) is negative - the movement is opposite to the sign inside the bracket.
What is the difference between y = pf(x) and y = f(px)?
\(y=pf(x)\) stretches the graph vertically by scale factor \(p\). \(y=f(px)\) stretches it horizontally by scale factor \(\tfrac1p\) - multiplying \(x\) inside the brackets compresses the graph when \(p\) is greater than 1.