Question

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

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

image of matching lines and vertex options: (0, -1), (-1, 0), (3, 0), (0, -3), (-3, 0), (3, 0), (0, 3), (-3, 2), (3, 2), (2, -3)

Explanation:

Step1: Recall vertex form of absolute value function

The general form of an absolute value function is \( f(x) = |x - h| + k \), where the vertex is \((h, k)\).

Step2: Find vertex for \( f(x)=|x - 3| \)

For \( f(x)=|x - 3| \), we can write it as \( f(x)=|x - 3|+0 \). Comparing with \( f(x) = |x - h| + k \), we have \( h = 3 \) and \( k = 0 \). So the vertex is \((3, 0)\).

Step3: Find vertex for \( f(x)=|x + 3|+2 \)

Rewrite \( f(x)=|x + 3|+2 \) as \( f(x)=|x - (- 3)|+2 \). Here, \( h=-3 \) and \( k = 2 \). So the vertex is \((-3, 2)\).

Step4: Find vertex for \( f(x)=|x|-3 \)

Rewrite \( f(x)=|x|-3 \) as \( f(x)=|x - 0|-3 \). Here, \( h = 0 \) and \( k=-3 \). So the vertex is \((0, - 3)\).

Step5: Find vertex for \( f(x)=|x + 1| \)

Rewrite \( f(x)=|x + 1| \) as \( f(x)=|x - (-1)|+0 \). Here, \( h=-1 \) and \( k = 0 \). So the vertex is \((-1, 0)\).

Answer:

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