Question

name: josue arnuldo nunha
date:
graphing absolute value
complete the following tables.
1).
$f(x) = |x - 2|$
2).
$f(x) = 2|x + 2|$
3).
$f(x) = -2|x + 1|$
4).
$f(x) = |x^2 - 2x|$
5).
$f(x) = 3|2x - 10| + 5$
(each function corresponds to a table with columns $x$ and $y$ to be filled)

Explanation:

Step1: Calculate for $f(x)=|x-2|$ at $x=3$

Substitute $x=3$ into the function:
$y = |3 - 2| = |1| = 1$

Step2: Choose $x$-values for $f(x)=2|x+2|$

Pick vertex $x=-2$, and $x=0, -4$:

• At $x=-2$: $y=2|-2+2|=2|0|=0$
• At $x=0$: $y=2|0+2|=2|2|=4$
• At $x=-4$: $y=2|-4+2|=2|-2|=4$

Step3: Choose $x$-values for $f(x)=-2|x+1|$

Pick vertex $x=-1$, and $x=0, -2$:

• At $x=-1$: $y=-2|-1+1|=-2|0|=0$
• At $x=0$: $y=-2|0+1|=-2|1|=-2$
• At $x=-2$: $y=-2|-2+1|=-2|-1|=-2$

Step4: Choose $x$-values for $f(x)=|x^2-2x|$

Pick $x=0, 1, 2, 3$:

• At $x=0$: $y=|0^2-2(0)|=|0|=0$
• At $x=1$: $y=|1^2-2(1)|=|1-2|=|-1|=1$
• At $x=2$: $y=|2^2-2(2)|=|4-4|=0$
• At $x=3$: $y=|3^2-2(3)|=|9-6|=|3|=3$

Step5: Choose $x$-values for $f(x)=3|2x-10|+5$

Pick vertex $x=5$, and $x=0, 10$:

• At $x=5$: $y=3|2(5)-10|+5=3|10-10|+5=0+5=5$
• At $x=0$: $y=3|2(0)-10|+5=3|-10|+5=30+5=35$
• At $x=10$: $y=3|2(10)-10|+5=3|20-10|+5=30+5=35$

Answer:

1. For $f(x)=|x-2|$:

$x$	$y$
0	2
3	1

1. For $f(x)=2|x+2|$:

$x$	$y$
0	4
-4	4

1. For $f(x)=-2|x+1|$:

$x$	$y$
0	-2
-2	-2

1. For $f(x)=|x^2-2x|$:

$x$	$y$
1	1
2	0
3	3

1. For $f(x)=3|2x-10|+5$:

$x$	$y$
0	35
10	35