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example 4 write an equation involving absolute value for each graph. 24…

Question

example 4
write an equation involving absolute value for each graph.
24.

$$\begin{tikzpicture} \\draw-> (-5,0) -- (5,0); \\foreach \\x in {-5,-4,-3,-2,-1,0,1,2,3,4,5} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-2,0) circle (2pt); \\filldrawblue (4,0) circle (2pt); \\end{tikzpicture}$$

25.

$$\begin{tikzpicture} \\draw-> (-10,0) -- (10,0); \\foreach \\x in {-10,-8,-6,-4,-2,0,2,4,6,8,10} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-6,0) circle (2pt); \\filldrawblue (6,0) circle (2pt); \\end{tikzpicture}$$

26.

$$\begin{tikzpicture} \\draw-> (-5,0) -- (5,0); \\foreach \\x in {-5,-4,-3,-2,-1,0,1,2,3,4,5} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-3,0) circle (2pt); \\filldrawblue (5,0) circle (2pt); \\end{tikzpicture}$$

27.

$$\begin{tikzpicture} \\draw-> (-7,0) -- (3,0); \\foreach \\x in {-7,-6,-5,-4,-3,-2,-1,0,1,2,3} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-6,0) circle (2pt); \\filldrawblue (2,0) circle (2pt); \\end{tikzpicture}$$

28.

$$\begin{tikzpicture} \\draw-> (-5,0) -- (5,0); \\foreach \\x in {-5,-4,-3,-2,-1,0,1,2,3,4,5} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-1,0) circle (2pt); \\filldrawblue (1,0) circle (2pt); \\end{tikzpicture}$$

29.

$$\begin{tikzpicture} \\draw-> (-7,0) -- (3,0); \\foreach \\x in {-7,-6,-5,-4,-3,-2,-1,0,1,2,3} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-5,0) circle (2pt); \\filldrawblue (-1,0) circle (2pt); \\end{tikzpicture}$$

30.

$$\begin{tikzpicture} \\draw-> (-5,0) -- (5,0); \\foreach \\x in {-5,-4,-3,-2,-1,0,1,2,3,4,5} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (3,0) circle (2pt); \\filldrawblue (5,0) circle (2pt); \\end{tikzpicture}$$

31.

$$\begin{tikzpicture} \\draw-> (-5,0) -- (5,0); \\foreach \\x in {-5,-4,-3,-2,-1,0,1,2,3,4,5} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x}; \\filldrawblue (-4,0) circle (2pt); \\filldrawblue (4,0) circle (2pt); \\end{tikzpicture}$$

Explanation:

Response

Let's solve problem 24 as an example (we can solve others similarly).

Problem 24:
Step 1: Find the midpoint (average) of the two points.

The two points are at \( x = -2 \) and \( x = 4 \). The midpoint \( h \) is calculated as:
\[
h=\frac{-2 + 4}{2}=\frac{2}{2}=1
\]

Step 2: Find the distance from the midpoint to either point (this is the radius \( r \)).

Distance from \( 1 \) to \( 4 \) (or to \( -2 \)):
\[
r=\vert4 - 1\vert=\vert-2 - 1\vert = 3
\]

Step 3: Write the absolute - value equation.

The general form of an absolute - value equation for a number line graph with solutions \( x_1 \) and \( x_2 \) is \( \vert x - h\vert=r \), where \( h \) is the midpoint and \( r \) is the distance from the midpoint to each solution.
Substituting \( h = 1 \) and \( r = 3 \) into the formula, we get:
\[
\vert x - 1\vert=3
\]

Problem 25:
Step 1: Find the midpoint of the two points.

The two points are at \( x=-6 \) and \( x = 6 \). The midpoint \( h \) is:
\[
h=\frac{-6 + 6}{2}=0
\]

Step 2: Find the distance from the midpoint to either point.

Distance from \( 0 \) to \( 6 \) (or to \( -6 \)):
\[
r=\vert6-0\vert=\vert - 6-0\vert=6
\]

Step 3: Write the absolute - value equation.

Using the formula \( \vert x - h\vert=r \) with \( h = 0 \) and \( r = 6 \), we have:
\[
\vert x\vert=6
\]

Problem 26:
Step 1: Find the midpoint of the two points.

The two points are at \( x=-3 \) and \( x = 5 \). The midpoint \( h \) is:
\[
h=\frac{-3 + 5}{2}=\frac{2}{2}=1
\]

Step 2: Find the distance from the midpoint to either point.

Distance from \( 1 \) to \( 5 \) (or to \( -3 \)):
\[
r=\vert5 - 1\vert=\vert-3 - 1\vert = 4
\]

Step 3: Write the absolute - value equation.

Using the formula \( \vert x - h\vert=r \) with \( h = 1 \) and \( r = 4 \), we get:
\[
\vert x - 1\vert=4
\]

Problem 27:
Step 1: Find the midpoint of the two points.

The two points are at \( x=-6 \) and \( x = 2 \). The midpoint \( h \) is:
\[
h=\frac{-6+2}{2}=\frac{-4}{2}=-2
\]

Step 2: Find the distance from the midpoint to either point.

Distance from \( - 2 \) to \( 2 \) (or to \( -6 \)):
\[
r=\vert2-(-2)\vert=\vert-6 - (-2)\vert=\vert4\vert = 4
\]

Step 3: Write the absolute - value equation.

Using the formula \( \vert x - h\vert=r \) with \( h=-2 \) and \( r = 4 \), we have:
\[
\vert x+2\vert=4
\]

Problem 28:
Step 1: Find the midpoint of the two points.

The two points are at \( x=-1 \) and \( x = 1 \). The midpoint \( h \) is:
\[
h=\frac{-1 + 1}{2}=0
\]

Step 2: Find the distance from the midpoint to either point.

Distance from \( 0 \) to \( 1 \) (or to \( -1 \)):
\[
r=\vert1-0\vert=\vert-1 - 0\vert = 1
\]

Step 3: Write the absolute - value equation.

Using the formula \( \vert x - h\vert=r \) with \( h = 0 \) and \( r = 1 \), we get:
\[
\vert x\vert=1
\]

Problem 29:
Step 1: Find the midpoint of the two points.

The two points are at \( x=-5 \) and \( x=-1 \). The midpoint \( h \) is:
\[
h=\frac{-5+(-1)}{2}=\frac{-6}{2}=-3
\]

Step 2: Find the distance from the midpoint to either point.

Distance from \( -3 \) to \( -1 \) (or to \( -5 \)):
\[
r=\vert-1-(-3)\vert=\vert-5 - (-3)\vert=\vert2\vert = 2
\]

Step 3: Write the absolute - value equation.

Using the formula \( \vert x - h\vert=r \) with \( h=-3 \) and \( r = 2 \), we have:
\[
\vert x + 3\vert=2
\]

Problem 30:
Step 1: Find the midpoint of the two points.

The two points are at \( x = 3 \) and \( x = 5 \). The midpoint \( h \) is:
\[
h=\frac{3 + 5}{2}=\frac{8}{2}=4
\]

Step 2: Find the distance from the midpoint to either…

Answer:

s:

  1. \(\boldsymbol{\vert x - 1\vert=3}\)
  1. \(\boldsymbol{\vert x\vert=6}\)
  1. \(\boldsymbol{\vert x - 1\vert=4}\)
  1. \(\boldsymbol{\vert x + 2\vert=4}\)
  1. \(\boldsymbol{\vert x\vert=1}\)
  1. \(\boldsymbol{\vert x + 3\vert=2}\)
  1. \(\boldsymbol{\vert x - 4\vert=1}\)
  1. \(\boldsymbol{\vert x\vert=4}\)