# How do you graph and solve #5 + |1-x/2| >=8#?

##### 1 Answer

#### Explanation:

**1) Simplifying**

First of all, bring

#5 + abs(1 - x/2) >= 8#

#<=> abs(1 - x/2) >= 3 #

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**2) Evaluating the absolute value function**

To evaluate the absolute value function, we need to find out when

To do this, let's find the point where

# 1 - x/2 = 0 " " <=> " " x/2 = 1 " " <=> " " x = 2#

Plugging

# 1 - x/2 >= 0 color(white)(xxx) "for " x <= 2#

# 1 - x/2 < 0 color(white)(xxx) "for " x > 2#

Now, you can evaluate the absolute value function:

# abs(1 - x/2) = { (color(white)(xx) 1 - x/2, color(white)(xxx) "for " 1 - x/2 >= 0 ), (-(1 - x/2), color(white)(xxx) "for " 1 - x/2 < 0) :}#

# color(white)(xxxxx) = { (color(white)(x) 1 - x/2, color(white)(xxxxx) "for " x <= 2 ), (-1 + x/2, color(white)(xxxxx) "for " x > 2) :}#

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**3) Solving the two cases**

**3a)** Let

This means that

# => 1 - x/2 >= 3#

... subtract

#<=> - x/2 >= 2#

... multiply both sides with

Be careful: if multiplying with a negative number or dividing by a negative number, you need to flip the inequality sign!

#<=> x <= - 4#

Now, we need to combine the condition

As

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**3b)** Let

This means that

# => - 1 + x/2 >= 3#

... add

# <=> x/2 >= 4#

... multiply both sides of the inequality with

#<=> x >= 8#

Between the two conditions,

Thus, this is the solution for the second case.

In total, the solution is

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**4) Graphing**

You can graph the absolute value function

- the "elbow" is the point of the function where
#1 - x/2 = 0# holds which is#x = 2# . Thus, the elbow is#(2; 0)# . - The slope is the factor of
#x/2# , so it's#1/2# .

Thus the absolute function looks as follows:

graph{abs(1 - x/2) [-10, 10, -5, 5]}

The graph of

#abs(1 - x/2) >= 3#

is the part of the graph that is above the horizontal line at

graph{(y - abs(1 - x/2))(y - 3) = 0 [-15, 15, -5, 10]}

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Hope that this helped! :-)