Absolute value functions - TCE Mathematics Specialised (Tasmania)

Sketch $y = |2x - 4|$.

Step 1: Sketch the underlying straight line

Before applying the absolute value, sketch the ordinary line $y = 2x - 4$. This line has gradient $2$, crosses the $x$ axis at $x = 2$ (set $y = 0$), and crosses the $y$ axis at $y = -4$. Knowing where the line crosses zero is essential because that is exactly the point where the reflection must happen.

Step 2: Apply the reflection rule to get the modulus graph

The absolute value maps any output below zero to its positive mirror image. For $x \ge 2$ the line $2x - 4$ is zero or positive, so that part of the graph is unchanged. For $x < 2$ the line is negative, so reflect that portion upward in the $x$ axis. The result is a V-shape with its vertex at $(2, 0)$, gradient $2$ to the right and gradient $-2$ to the left. The original $y$ intercept of $-4$ becomes $|{-4}| = 4$ after reflection.

Final answer: A V-shape with vertex at $(2, 0)$, gradient $2$ for $x \ge 2$, gradient $-2$ for $x < 2$, and $y$ intercept at $4$.

Solve $|3x - 1| = 5$.

Step 1: Split into two cases

The definition of absolute value says $|A| = 5$ whenever $A = 5$ or $A = -5$, because both give $|A| = 5$. Applying this to the expression inside the modulus produces two separate linear equations that can each be solved independently:

Step 2: Check both solutions in the original equation

It is good practice to verify each answer, since occasionally an extraneous solution can appear. Substituting back confirms both values satisfy the original equation: $|3(2) - 1| = |5| = 5$ and $|3(-\tfrac43) - 1| = |-5| = 5$.

Final answer: $x = 2$ or $x = -\tfrac{4}{3}$.

Solve $|x + 2| \le 3$.

Step 1: Rewrite using the "less than" inequality rule

The inequality $|A| \le k$ (with $k > 0$) is equivalent to the compound inequality $-k \le A \le k$, because the absolute value is at most $k$ whenever the expression inside lies between $-k$ and $k$. Applying this standard pattern removes the modulus sign in one step and replaces it with a double inequality:

Step 2: Isolate $x$ by subtracting throughout

The expression inside the inequality is $x + 2$, so subtracting $2$ from all three parts isolates $x$ without changing the direction of either inequality sign:

Final answer: The solution set is the closed interval $[-5, 1]$, which is all real numbers within distance $3$ of $-2$ on the number line.

Solve $|2x - 1| = x + 3$.

Step 1: Identify the critical point and set up two cases

The expression inside the modulus, $2x - 1$, changes sign at $x = \tfrac12$. For $x \ge \tfrac12$ the expression is non-negative, so $|2x - 1| = 2x - 1$; for $x < \tfrac12$ it is negative, so $|2x - 1| = -(2x - 1)$. Splitting at this critical point means that within each region the modulus can be removed and the resulting equation is a straightforward linear one.

Step 2: Solve in each region and check validity

For $x \ge \tfrac12$, remove the modulus directly and solve:

Since $4 \ge \tfrac12$, this solution lies in the correct region and is valid.

For $x < \tfrac12$, the inside is negative so apply the negative case and solve:

Since $-\tfrac23 < \tfrac12$, this also lies in its correct region and is valid.

Step 3: Confirm both solutions

A final substitution back into $|2x - 1| = x + 3$ confirms there are no errors: $|2(4) - 1| = 7 = 4 + 3$, and $|2(-\tfrac23) - 1| = |-\tfrac73| = \tfrac73 = -\tfrac23 + 3$. The right side $x + 3$ is positive at both solutions, which is necessary because the left side is an absolute value.

Final answer: $x = 4$ or $x = -\tfrac23$.