The modulus functionEdexcel International A Level Maths: Revision notes

The modulus (absolute value) of a number is its distance from zero, so it is never negative: $|x|=\begin{cases}x,&x\geqslant0\\-x,&x<0\end{cases}$ The function $y=|ax+b|$ gives a V-shaped graph. The vertex is where $ax+b=0$, at $x=-\frac{b}{a}$, with $y=0$. For $y=|2x-1|$ the vertex is $\left(\frac12,0\right)$ and the $y$-intercept is $(0,1)$. Each side of the vertex is a straight line with gradient $\pm a$.

$y=|f(x)|$: keep every part of the graph of $y=f(x)$ that lies on or above the $x$-axis, and reflect the parts below it in the $x$-axis. For $f(x)=x^2-4x+3$ the minimum $(2,-1)$ becomes a local maximum $(2,1)$. $y=f(|x|)$: keep the graph for $x\geqslant0$, discard the part for $x<0$, and reflect the $x\geqslant0$ part in the $y$-axis. The graph is symmetrical about the $y$-axis. For $p(x)=x^2-x-6$, $p(|x|)$ has $x$-intercepts $\pm3$ and $y$-intercept $-6$.

To solve $|f(x)|=g(x)$, split into two cases: $f(x)=g(x)$ and $f(x)=-g(x)$. Then check each answer, because $g(x)$ must be non-negative. Example: $|2x-1|=x+5$. Case 1: $2x-1=x+5$, so $x=6$. Case 2: $2x-1=-(x+5)$, so $x=-\frac43$. Both give a positive right-hand side, so both are valid. Alternatively square both sides: $(2x-1)^2=(x+5)^2$ gives $3x^2-14x-24=0$, with the same roots, but still check them in the original equation.

Find the boundary values by solving the equation, then use a sketch or a test value to decide which regions satisfy the inequality. For $|2x-1|>x+5$ the boundaries are $x=-\frac43$ and $x=6$. The V lies above the line outside them, so $x<-\frac43$ or $x>6$. For $|3x-2|<2x+1$ the boundaries are $\frac15$ and $3$ and the V lies below the line between them, so $\frac15<x<3$. Squaring both sides is also valid for $|a|<|b|$, giving $a^2<b^2$.

For $|f(x)|=k$, imagine the horizontal line $y=k$ crossing the graph of $y=|f(x)|$. With $p(x)=x^2-x-6$, the local maximum of $|p(x)|$ is $\frac{25}{4}$, so $|p(x)|=k$ has exactly four solutions when $0<k<\frac{25}{4}$. Always give the coordinates of vertices and intercepts, and show both cases when solving.