Question

identify the vertex. determine if the graph opens up or down (circle). determine if the graph has a maximum or minimum (circle) and its value. decide if the graph is narrower, wider, or the same width as the parent graph (circle)

1. (y = -|x + 1|) vertex ((-1), (0)) opens up / down maximum minimum of (-1) narrower / wider / same
2. (y = 7|x - 3|-4) vertex (__, ) opens up / down maximum minimum of __ narrower / wider / same
3. (y=-\frac{1}{3}|x - 1|) vertex (__, ) opens up / down maximum minimum of __ narrower / wider / same
4. (y=\frac{3}{2}|x + 9|-1) vertex (__, ) opens up / down maximum minimum of __ narrower / wider / same
5. (y=\frac{1}{4}|x + 3|-6) vertex (__, ) opens up / down maximum minimum of __ narrower / wider / same
6. (y=-|x| + 5) vertex (__, ) opens up / down maximum minimum of __ narrower / wider / same

neatly graph each absolute - value function.

1. (y = 3|x - 3|)
2. (y=-|x| + 4)
3. (y=-|x + 3|+5)
4. (y = 2|x + 1|-1)
5. (y=\frac{1}{3}|x + 2|-5)
6. (y=-\frac{3}{2}|x - 3|+2)

Explanation:

Step1: Recall vertex - form of absolute - value function

The general form of an absolute - value function is $y=a|x - h|+k$, and its vertex is $(h,k)$.

Step2: Determine the vertex for $y = -|x + 1|$

For $y=-|x + 1|= - 1|x-(-1)|+0$, the vertex is $(-1,0)$. Since $a=-1<0$, the graph opens down. The maximum value occurs at the vertex, and the maximum value is $y = 0$. The coefficient $|a| = 1$, compared to the parent function $y = |x|$ (where $a = 1$), the width is the same.

Step3: Determine the vertex for $y = 7|x - 3|-4$

For $y = 7|x - 3|-4$, the vertex is $(3,-4)$. Since $a = 7>0$, the graph opens up. The minimum value occurs at the vertex, and the minimum value is $y=-4$. Since $|a|=7>1$ (compared to the parent function $y = |x|$), the graph is narrower.

Step4: Determine the vertex for $y=-\frac{1}{3}|x - 1|$

For $y=-\frac{1}{3}|x - 1|=-\frac{1}{3}|x - 1|+0$, the vertex is $(1,0)$. Since $a=-\frac{1}{3}<0$, the graph opens down. The maximum value occurs at the vertex, and the maximum value is $y = 0$. Since $|a|=\frac{1}{3}<1$ (compared to the parent function $y = |x|$), the graph is wider.

Step5: Determine the vertex for $y=\frac{3}{2}|x + 9|-1$

For $y=\frac{3}{2}|x + 9|-1$, the vertex is $(-9,-1)$. Since $a=\frac{3}{2}>0$, the graph opens up. The minimum value occurs at the vertex, and the minimum value is $y=-1$. Since $|a|=\frac{3}{2}>1$ (compared to the parent function $y = |x|$), the graph is narrower.

Step6: Determine the vertex for $y=\frac{1}{4}|x + 3|-6$

For $y=\frac{1}{4}|x + 3|-6$, the vertex is $(-3,-6)$. Since $a=\frac{1}{4}>0$, the graph opens up. The minimum value occurs at the vertex, and the minimum value is $y=-6$. Since $|a|=\frac{1}{4}<1$ (compared to the parent function $y = |x|$), the graph is wider.

Step7: Determine the vertex for $y=-|x|+5$

For $y=-|x|+5=-1|x - 0|+5$, the vertex is $(0,5)$. Since $a=-1<0$, the graph opens down. The maximum value occurs at the vertex, and the maximum value is $y = 5$. Since $|a| = 1$ (compared to the parent function $y = |x|$), the width is the same.

Answer:

1. Vertex: $(-1,0)$; Opens: DOWN; MAXIMUM of $0$; SAME
2. Vertex: $(3,-4)$; Opens: UP; MINIMUM of $-4$; NARROWER
3. Vertex: $(1,0)$; Opens: DOWN; MAXIMUM of $0$; WIDER
4. Vertex: $(-9,-1)$; Opens: UP; MINIMUM of $-1$; NARROWER
5. Vertex: $(-3,-6)$; Opens: UP; MINIMUM of $-6$; WIDER
6. Vertex: $(0,5)$; Opens: DOWN; MAXIMUM of $5$; SAME