The modulus (absolute value) function
Master the modulus (absolute value) function in Year 11 VCE Specialist Mathematics. 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. It sits in the Functions, relations and graphs area of study of the VCE Mathematics Study Design (VCAA), within the Functions, relations and graphs topic of Unit 2.
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 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.