Modulus Transformations (DP IB Analysis & Approaches (AA): HL): Revision Note

Modulus transformations

How do I sketch the modulus of a function y = |f(x)|?

• To sketch $y=\left|f\left(x\right)\right|$

  • STEP 1
    Sketch the parts of $y=f\left(x\right)$ that are on or above the $x$-axis

  • STEP 2
    Sketch the reflections in the $x$-axis of any parts below

How do I sketch a function of mod x, y = f(|x|)?

• To sketch $y=f\left(\left|x\right|\right)$

  • STEP 1
    Keep the part of $y=f\left(x\right)$ on, or to the right of, the $y$-axis

    • get rid of any part that is to the left

  • STEP 2

    Reflect the right-hand side in the $y$-axis to create a left-hand side

    • This will create a symmetric graph about the $y$-axis

What are some differences between y = |f(x)| or y = f(|x|)?

• It helps to remember that

  • $y=\left|f\left(x\right)\right|$ can never go below the $x$-axis

    • whereas $y=f\left(\left|x\right|\right)$ can

  • $y=f\left(\left|x\right|\right)$ is always symmetric about the $y$-axis

    • whereas $y=\left|f\left(x\right)\right|$ does not have to have any lines of symmetry

How do I sketch transformations of y = |f(x)|?

• To sketch $y=\left|af\left(x\right)+b\right|$

  • first sketch $y=f\left(x\right)$ without modulus signs

  • then apply a vertical stretch of scale factor $a$

    • $y=af\left(x\right)$

  • followed by a vertical translation of $b$

    • $y=af\left(x\right)+b$

  • then take the modulus (using the rules above)

    • $y=\left|af\left(x\right)+b\right|$

• To sketch $y=a\left|f\left(x\right)\right|+b$

  • sketch $y=\left|f\left(x\right)\right|$ (as above)

  • then apply a vertical stretch of scale factor $a$

    • $y=a\left|f\left(x\right)\right|$

  • followed by a vertical translation of $b$

    • $y=a\left|f\left(x\right)\right|+b$

How do I sketch transformations of y = f(|x|)?

• To sketch $y=af\left(\left|x\right|\right)+b$

  • sketch $y=f\left(\left|x\right|\right)$ (using the rules above)

  • then apply a vertical stretch of scale factor $a$

    • $y=af\left(\left|x\right|\right)$

  • followed by a vertical translation of $b$

    • $y=af\left(\left|x\right|\right)+b$

How do I sketch y = |f(ax+b)| or y = f(|ax+b|)?

• To sketch $y=\left|f\left(ax+b\right)\right|$

  • first sketch $y=f\left(ax+b\right)$ which factorises to $y=f\left(a\left(x+\frac{b}{a}\right)\right)$

    • i.e. first apply a horizontal stretch to $y=f\left(x\right)$ of scale factor $\frac{1}{a}$

    • $y=f\left(x\right)$ becomes $y=f\left(ax\right)$

    • followed by a horizontal translation of $\frac{b}{a}$ to the left, $\left(\begin{array}{c}-\frac{b}{a}\\ 0\end{array}\right)$

    • i.e. $y=f\left(ax\right)$ becomes $y=f\left(a\left(x+\frac{b}{a}\right)\right)$ which is $y=f\left(ax+b\right)$

  • then take the modulus of $y=f\left(ax+b\right)$ using the rules above

    • $y=\left|f\left(ax+b\right)\right|$

• To sketch $y=f\left(|ax+b|\right)$

  • factorise it to $f\left(\left|a\right|\left|x+\frac{b}{a}\right|\right)$

    • i.e. first apply a horizontal stretch of scale factor $\frac{1}{\left|a\right|}$ to $y=f\left(\left|x\right|\right)$

    • $y=f\left(\left|x\right|\right)$ becomes $y=f\left(\left|a\right|\left|x\right|\right)$

    • followed by a horizontal translation of $\frac{b}{a}$ to the left, $\left(\begin{array}{c}-\frac{b}{a}\\ 0\end{array}\right)$

    • i.e. $y=f\left(\left|a\right|\left|x\right|\right)$ becomes $f\left(\left|a\right|\left|x+\frac{b}{a}\right|\right)$ which is $y=f\left(|ax+b|\right)$