Question

u4: absolute value fct exit ticket
match the vertex to each absolute value function.

1. $f(x) = |x + 3| + 3$ 10) $f(x) = |x| - 4$ 11) $f(x) = |x + 1| - 4$ 12) $f(x) = |x - 2| + 2$

chart for matching problems 9 - 12 to vertices, with vertex options: (0,3), (-3,2), (3,2), (2, -3), (2,2), (-2, 2), (-1,4), (-1, -4), (-4, -1), (3,3), (3, -3), (-3, 3), (4,0), (-4,0), (0, -4)

Explanation:

Step1: Recall Vertex Form of Absolute Value Function

The general form of an absolute value function is \( f(x) = |x - h| + k \), where \((h, k)\) is the vertex. If the function is \( f(x) = |x + a| + b \), we can rewrite it as \( f(x) = |x - (-a)| + b \), so the vertex is \((-a, b)\).

Step2: Find Vertex for \( f(x) = |x + 3| + 3 \)

For \( f(x) = |x + 3| + 3 \), rewrite as \( f(x) = |x - (-3)| + 3 \). So \( h = -3 \), \( k = 3 \). Vertex is \((-3, 3)\).

Step3: Find Vertex for \( f(x) = |x| - 4 \)

Rewrite \( f(x) = |x| - 4 \) as \( f(x) = |x - 0| + (-4) \). So \( h = 0 \), \( k = -4 \). Vertex is \((0, -4)\).

Step4: Find Vertex for \( f(x) = |x + 1| - 4 \)

Rewrite as \( f(x) = |x - (-1)| + (-4) \). So \( h = -1 \), \( k = -4 \). Vertex is \((-1, -4)\).

Step5: Find Vertex for \( f(x) = |x - 2| + 2 \)

Rewrite as \( f(x) = |x - 2| + 2 \). So \( h = 2 \), \( k = 2 \). Vertex is \((2, 2)\).

Answer:

1. \( f(x) = |x + 3| + 3 \) matches with \((-3, 3)\)
2. \( f(x) = |x| - 4 \) matches with \((0, -4)\)
3. \( f(x) = |x + 1| - 4 \) matches with \((-1, -4)\)
4. \( f(x) = |x - 2| + 2 \) matches with \((2, 2)\)