\( f(x) = |x| \)

• Graph shape: V-shaped, vertex at origin (0, 0)

• Domain: All real numbers → \( (-\infty, \infty) \)

• Range: All non-negative real numbers → \( [0, \infty) \)

\( f(x) = a|x - h| + k \)

where:

• a = vertical stretch/compression and reflection

• h = horizontal shift (vertex x-coordinate)

• k = vertical shift (vertex y-coordinate)

• Vertex: (h, k)

Domain:

• Look at all x-values the graph covers (usually all real numbers)

• Unless restricted by context, domain = \( \mathbb{R} \) or \( (-\infty, \infty) \)

Range:

• Identify the vertex (turning point)

• If a > 0 (opens upward): Range = \( [k, \infty) \)

• If a < 0 (opens downward): Range = \( (-\infty, k] \)

Rule 1: No denominator or even root

→ Domain is all real numbers: \( \mathbb{R} \) or \( (-\infty, \infty) \)

Rule 2: Denominator present

→ Exclude values that make denominator = 0

Rule 3: Even root present (square root, etc.)

→ Exclude values that make radicand negative

For \( f(x) = a|x - h| + k \):

Step 1: Identify the vertex (h, k)

Step 2: Determine if \( a > 0 \) or \( a < 0 \)

Step 3: Write range:

• If \( a > 0 \): Range = \( [k, \infty) \)

• If \( a < 0 \): Range = \( (-\infty, k] \)

Example 1: \( f(x) = |x - 3| \)

Domain: \( \mathbb{R} \) (all real numbers)

Range: \( [0, \infty) \)

Example 2: \( f(x) = -2|x + 1| + 5 \)

Domain: \( \mathbb{R} \)

Vertex: (-1, 5), opens downward (a = -2 < 0)

Range: \( (-\infty, 5] \)

Example 3: \( f(x) = \frac{|x - 4|}{x - 4} \)

Denominator = 0 when x = 4

Domain: \( \mathbb{R} - \{4\} \) or \( (-\infty, 4) \cup (4, \infty) \)

Range: \( \{-1, 1\} \)

\( f(x) = a|x - h| + k \)

Step 1: Find the vertex (h, k)

• The vertex is at point (h, k)

Step 2: Determine if graph opens up or down

• If a > 0: opens upward (V shape)

• If a < 0: opens downward (∧ shape)

Step 3: Find the slope

• Right side slope = a

• Left side slope = -a

Step 4: Plot additional points

• From vertex, move right 1 unit, then up |a| units

• From vertex, move left 1 unit, then up |a| units

Step 5: Draw the V-shape

• Connect the points with straight lines forming a V

1. Choose x-values on both sides of h (the vertex x-coordinate)

2. Calculate corresponding y-values

3. Plot all points and connect with V-shape

\( f(x) = a|x - h| + k \)

Starting from parent function \( f(x) = |x| \)

⚠️ Sign Convention (Counter-intuitive!):

• \( |x - 3| \) means shift RIGHT 3 (subtract → right)

• \( |x + 3| \) means shift LEFT 3 (add → left)

• \( |x| + 3 \) means shift UP 3 (add → up)

• \( |x| - 3 \) means shift DOWN 3 (subtract → down)

Example 1: \( f(x) = 2|x - 3| + 1 \)

• Vertical stretch by factor of 2 (steeper)

• Horizontal shift right 3 units

• Vertical shift up 1 unit

• Vertex: (3, 1)

Example 2: \( f(x) = -\frac{1}{2}|x + 4| - 2 \)

• Reflection over x-axis (opens downward)

• Vertical compression by factor of 1/2 (wider)

• Horizontal shift left 4 units

• Vertical shift down 2 units

• Vertex: (-4, -2)

\( y < a|x - h| + k \) or \( y > a|x - h| + k \)

\( y \leq a|x - h| + k \) or \( y \geq a|x - h| + k \)

Step 1: Graph the Boundary Line

• Replace inequality symbol with = sign

• Graph the absolute value function (V-shape)

Step 2: Determine Line Type

• Use solid line for \( \leq \) or \( \geq \) (inclusive)

• Use dashed line for < or > (exclusive)

Step 3: Test a Point

• Choose a test point not on the boundary (often use origin if not on line)

• Substitute into original inequality to check if true

Step 4: Shade the Region

• If test point satisfies inequality → shade that region

• If test point doesn't satisfy → shade the opposite region

Step 1: Graph Each Inequality Separately

• Graph boundary line (solid or dashed)

• Shade appropriate region for each inequality

Step 2: Identify the Intersection Region

• Look for where ALL shaded regions overlap

• This overlap is the solution set

Step 3: Indicate the Solution

• Use different shading or color for the final solution region

• Label clearly or use darker shading for overlap

Step 4: Verify with a Test Point

• Choose a point in the solution region

• Substitute into ALL original inequalities

• Point should satisfy ALL inequalities

System:

\( y \geq |x - 2| + 1 \)

\( y < 2x + 3 \)

Solution Process:

1. Graph \( y = |x - 2| + 1 \) with solid line (≥), shade outside/above V

2. Graph \( y = 2x + 3 \) with dashed line (<), shade below line

3. Solution = region where both shadings overlap

No Solution:

• If shaded regions don't overlap, system has no solution

Unbounded Solution:

• If overlap region extends infinitely, solution is unbounded

Bounded Solution:

• If overlap creates a closed region, solution is bounded