Sketching the Graph of a Modulus Function

To sketch $y = |2x - 4|$, first picture the line inside the modulus, then reflect anything below the $x$-axis upward. The result is a V-shape with its vertex where the inside equals zero.

Sketch the graph of $y = |2x - 4|$, showing the coordinates of the vertex and the axis intercepts, and state the range. [4]

The working

Step 1, find the vertex where the inside is zero: $2x - 4 = 0 \implies x = 2, \quad \text{so the vertex is } (2, 0). \quad \text{(M1)}$

Step 2, find the intercepts:

• $y$-intercept ($x = 0$): $\ y = |{-4}| = 4$, giving $(0, 4)$.
• $x$-intercept: the vertex itself, $(2, 0)$. (A1)

Step 3, describe the two arms: $y = \begin{cases} 2x - 4, & x \ge 2 \ (\text{gradient } +2)\\ 4 - 2x, & x < 2 \ (\text{gradient } -2)\end{cases}$ so the graph is a symmetric V opening upward. (B1)

Step 4, state the range: since $|{\cdots}| \ge 0$, $y \ge 0. \quad \text{(B1)}$

Where the marks are won and lost

• The vertex sits on the $x$-axis at $x = 2$, because the modulus can’t be negative. Marks are lost when the V is drawn dipping below the axis.
• Label the $y$-intercept $(0,4)$: substitute $x=0$ into the modulus, giving $|-4| = 4$.
• The two arms are mirror images, with gradients $+2$ and $-2$.

Common mistakes

• Leaving the graph below the $x$-axis (drawing the plain line $y = 2x - 4$).
• Putting the vertex at $(0,4)$ instead of $(2,0)$.
• Stating the range as “all real $y$” rather than $y \ge 0$.

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