IB Mathematics AA SL Transformations of graphs Study Notes
IB Mathematics AA SL Transformations of graphs Study Notes Offer a clear explanation of Transformations of graphs , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Transformations of graphs
Transformation of Graphs
Transformation of graphs involves various operations that change the position or shape of a graph. Common transformations include translations, reflections, and stretches.
Guidance, Clarification, and Syllabus Links:
- Translations:
- Vertical translation: \( y = f(x) + b \), moves the graph up by \( b \) units if \( b > 0 \) or down by \( b \) units if \( b < 0 \).
- Horizontal translation: \( y = f(x – a) \), moves the graph to the right by \( a \) units if \( a > 0 \) or to the left by \( a < 0 \).
- Reflections:
- Reflection in the x-axis: \( y = -f(x) \), flips the graph over the x-axis.
- Reflection in the y-axis: \( y = f(-x) \), flips the graph over the y-axis.
- Vertical Stretch:
- With scale factor \( p \): \( y = pf(x) \). If \( p > 1 \), the graph stretches away from the x-axis. If \( 0 < p < 1 \), the graph compresses towards the x-axis.
- Horizontal Stretch:
- With scale factor \( \frac{1}{q} \): \( y = f(qx) \). If \( q > 1 \), the graph compresses towards the y-axis. If \( 0 < q < 1 \), the graph stretches away from the y-axis.
- Composite Transformations:
- Order matters when performing multiple transformations. For example, applying a vertical stretch first and then translating will yield a different graph than performing the translation first.
Examples of Transformations
Example 1: Translation of a Graph
Given the function \( y = x^2 \), apply the following transformations:
- Translation up by 3 units: \( y = x^2 + 3 \)
Example 2: Reflection of a Graph
Given the function \( y = x^3 \), reflect it over the x-axis and then the y-axis:
- Reflection over the x-axis: \( y = -x^3 \)
IB Mathematics AA SL Transformation of Graphs Exam Style Worked Out Questions
Question
The following diagram shows the graph of a function \(f\), for −4 ≤ x ≤ 2.
On the same axes, sketch the graph of \(f\left( { – x} \right)\).
Another function, \(g\), can be written in the form \(g\left( x \right) = a \times f\left( {x + b} \right)\). The following diagram shows the graph of \(g\).
Write down the value of a and of b.
▶️Answer/Explanation
Markscheme
A2 N2
[2 marks]
recognizing horizontal shift/translation of 1 unit (M1)
eg b = 1, moved 1 right
recognizing vertical stretch/dilation with scale factor 2 (M1)
eg a = 2, y ×(−2)
a = −2, b = −1 A1A1 N2N2
[4 marks]
Question
Let \(f(x) = 3{(x + 1)^2} – 12\) .
Show that \(f(x) = 3{x^2} + 6x – 9\) .[2]
For the graph of f
(i) write down the coordinates of the vertex;
(ii) write down the equation of the axis of symmetry;
(iii) write down the y-intercept;
(iv) find both x-intercepts.[8]
Hence sketch the graph of f .[2]
Let \(g(x) = {x^2}\) . The graph of f may be obtained from the graph of g by the two transformations:
a stretch of scale factor t in the y-direction
followed by a translation of \(\left( {\begin{array}{*{20}{c}}
p\\
q
\end{array}} \right)\) .
Find \(\left( {\begin{array}{*{20}{c}}
p\\
q
\end{array}} \right)\) and the value of t.[3]
▶️Answer/Explanation
Markscheme
\(f(x) = 3({x^2} + 2x + 1) – 12\) A1
\( = 3{x^2} + 6x + 3 – 12\) A1
\( = 3{x^2} + 6x – 9\) AG N0
[2 marks]
(i) vertex is \(( – 1{\text{, }} – 12)\) A1A1 N2
(ii) \(x = – 1\) (must be an equation) A1 N1
(iii) \((0{\text{, }} – 9)\) A1 N1
(iv) evidence of solving \(f(x) = 0\) (M1)
e.g. factorizing, formula,
correct working A1
e.g. \(3(x + 3)(x – 1) = 0\) , \(x = \frac{{ – 6 \pm \sqrt {36 + 108} }}{6}\)
\(( – 3{\text{, }}0)\), \((1{\text{, }}0)\) A1A1 N1N1
[8 marks]
A1A1 N2
Note: Award A1 for a parabola opening upward, A1 for vertex and intercepts in approximately correct positions.
[2 marks]
\(\left( {\begin{array}{*{20}{c}}
p\\
q
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
{ – 12}
\end{array}} \right)\) , \(t = 3\) (accept \(p = – 1\) , \(q = – 12\) , \(t = 3\) ) A1A1A1 N3
[3 marks]