The modulus function
Get on top of the modulus function for Year 11 Specialist Mathematics in Queensland (QCAA). The absolute value \(|x|\) measures how far a number sits from zero, and its graph is the tidy V-shape \(y=|x|\) that sits at the heart of this sketching-graphs topic.
You will learn to read the vertex and intercepts of \(y=a|x-h|+k\), solve modulus equations such as \(|ax+b|=c\), and build the graphs of \(y=|f(x)|\) and \(y=f(|x|)\) by reflection — skills you will reuse across functions and calculus later in the course.
Theory
The modulus function \(y=|x|\) turns every input into its distance from zero, giving a V-shaped graph. In Year 11 Specialist Mathematics (QCAA, Queensland) you read the vertex and intercepts of \(y=a|x-h|+k\), solve equations such as \(|ax+b|=c\), and build the graphs of \(y=|f(x)|\) and \(y=f(|x|)\) from \(y=f(x)\).
The absolute value (or modulus) of a number is its distance from zero on the number line, so it is never negative: \(|4|=4\) and \(|-4|=4\). As a rule, \(|x|\) leaves a non-negative input unchanged and flips the sign of a negative input.
Written as a piecewise function, \(|x|=x\) for \(x\ge 0\) and \(|x|=-x\) for \(x<0\). The graph of \(y=|x|\) is a V-shape with its point (the vertex) at the origin, made of two straight rays that are symmetric about the \(y\)-axis.
The general modulus graph \(y=a|x-h|+k\) is that same V translated to a vertex at \((h,k)\); \(a\) stretches it and, if \(a<0\), flips it to open downward. Setting \(y=0\) or \(x=0\) gives the \(x\)- and \(y\)-intercepts.
Two related graphs come from any \(y=f(x)\). For \(y=|f(x)|\) you reflect every part of the graph that is below the \(x\)-axis up above it. For \(y=f(|x|)\) you keep the right-hand side (\(x\ge 0\)) and mirror it across the \(y\)-axis, making an even function.
The piecewise definition of the absolute value:
The transformed V-graph and its vertex:
Solving a modulus equation splits into two cases:
How to work with a modulus graph or equation
- Read the vertex. For \(y=a|x-h|+k\) the V turns where the inside is zero, at \((h,k)\); the sign of \(a\) tells you whether it opens up or down.
- Find the intercepts. Put \(x=0\) for the \(y\)-intercept; set \(y=0\) and solve the resulting \(|\ldots|=\) number for the \(x\)-intercepts.
- Solve equations by cases. Replace \(|ax+b|=c\) with \(ax+b=c\) and \(ax+b=-c\); for \(|f(x)|=g(x)\) split on the sign of \(f(x)\) and check each answer fits its branch.
- Transform by reflection. For \(y=|f(x)|\) flip the below-axis part up; for \(y=f(|x|)\) mirror the \(x\ge 0\) side across the \(y\)-axis.
Read the vertex from \(y=a|x-h|+k\) (the V turns where the inside is zero):
| \(\text{inside}=0\) | \(\Rightarrow\) | \(x-1=0\) |
| \(x\) | \(=\) | \(1\) |
| \(\text{vertex}\) | \(=\) | \((1,-4)\) |
Put \(x=0\) for the \(y\)-intercept:
| \(y\) | \(=\) | \(2|0-1|-4\) |
| \(=\) | \(2(1)-4\) | |
| \(=\) | \(-2\) |
Set \(y=0\) and solve the modulus for the \(x\)-intercepts:
| \(2|x-1|-4\) | \(=\) | \(0\) |
| \(|x-1|\) | \(=\) | \(2\) |
| \(x-1\) | \(=\) | \(2 \;\text{ or }\; x-1=-2\) |
| \(x\) | \(=\) | \(3 \;\text{ or }\; x=-1\) |
Vertex \((1,-4)\); \(y\)-intercept \((0,-2)\); \(x\)-intercepts \((-1,0)\) and \((3,0)\).
The modulus equals a positive number, so split into two cases:
| \(3x+2\) | \(=\) | \(8\) |
| \(3x\) | \(=\) | \(6\) |
| \(x\) | \(=\) | \(2\) |
Now the negative case, \(3x+2=-8\):
| \(3x+2\) | \(=\) | \(-8\) |
| \(3x\) | \(=\) | \(-10\) |
| \(x\) | \(=\) | \(-\dfrac{10}{3}\) |
\(x=2\) or \(x=-\dfrac{10}{3}\).
Where the line is on or above the axis (\(x\ge 3\)) it is unchanged; below the axis (\(x<3\)) the sign flips:
| \(x\ge 3:\; |x-3|\) | \(=\) | \(x-3\) |
| \(x<3:\; |x-3|\) | \(=\) | \(-(x-3)\) |
| \(=\) | \(3-x\) |
The join (vertex) is where \(x-3=0\); read the \(y\)-intercept at \(x=0\):
| \(\text{vertex}\) | \(=\) | \((3,0)\) |
| \(y(0)\) | \(=\) | \(|0-3|\) |
| \(=\) | \(3\) |
The V has vertex \((3,0)\) and \(y\)-intercept \((0,3)\); the below-axis part of the line is reflected up.
Split on the sign of the inside. First the branch \(x\ge 4\), where \(|x-4|=x-4\):
| \(x-4\) | \(=\) | \(x-2\) |
| \(-4\) | \(=\) | \(-2 \;\;(\text{false})\) |
So there is no solution with \(x\ge 4\). Now the branch \(x<4\), where \(|x-4|=4-x\):
| \(4-x\) | \(=\) | \(x-2\) |
| \(6\) | \(=\) | \(2x\) |
| \(x\) | \(=\) | \(3\) |
Check \(x=3\) fits its branch and the original equation:
| \(x=3\) | \(<\) | \(4 \;\;(\text{valid})\) |
| \(|3-4|\) | \(=\) | \(1 \;=\; 3-2 \;\checkmark\) |
\(x=3\) (the only solution).
Common pitfalls
Frequently asked questions
What does the modulus (absolute value) of a number mean?
It is the distance of the number from zero, so it is never negative: \(|4|=4\) and \(|-4|=4\).
Where is the vertex of \(y=a|x-h|+k\)?
At \((h,k)\). The V turns where the inside \(x-h=0\), and \(k\) is the height there; \(a\) controls how steep it is and whether it opens up (\(a>0\)) or down (\(a<0\)).
How do you solve an equation like \(|3x+2|=8\)?
Split it into \(3x+2=8\) and \(3x+2=-8\), then solve each linear equation. Here \(x=2\) or \(x=-\tfrac{10}{3}\).
How many solutions does \(|ax+b|=c\) have?
Two when \(c>0\), one when \(c=0\), and none when \(c<0\), because a modulus is never negative.
What is the difference between \(y=|f(x)|\) and \(y=f(|x|)\)?
For \(y=|f(x)|\) you reflect the below-axis part of the graph up. For \(y=f(|x|)\) you keep the right-hand side and mirror it across the \(y\)-axis, making an even function.
Why is \(y=f(|x|)\) always symmetric about the \(y\)-axis?
Because replacing \(x\) with \(|x|\) gives the same output for \(x\) and \(-x\), so \(g(-x)=g(x)\); that is exactly the definition of an even function.