
a.
Tabular : Express each statement using an inequality involving absolute value. Do not solve. Copy and complete the table .Substitute the x and f(x) values for each point into each inequality. Mark whether the resulting statement is true or false.
a.

Answer to Problem 42PPS
Point | f(x)≥x−1 . | True/false | f(x)≤x−1 . | True/false |
(-4,2) | 2≥−4−1 | true | 2≤−4−1 | false |
(-2,2) | 2≥−2−1 | true | 2≤−2−1 | false |
(0.2) | 2≥−0−1 | true | 2≤0−1 | false |
(2,2) | 2≥2−1 | true | 2≤2−1 | false |
(4,2) | 2≥4−1 | false | 2≤3−1 | true |
Explanation of Solution
Given: Given table is given below.
Calculation:
Fill in the table by substituting each point into the given inequalities determining if the inequalities are true or false.
Point | f(x)≥x−1 . | True/false | f(x)≤x−1 . | True/false |
(-4,2) | 2≥−4−1 | true | 2≤−4−1 | false |
(-2,2) | 2≥−2−1 | true | 2≤−2−1 | false |
(0.2) | 2≥−0−1 | true | 2≤0−1 | false |
(2,2) | 2≥2−1 | true | 2≤2−1 | false |
(4,2) | 2≥4−1 | false | 2≤3−1 | true |
b.
You will investigate the graphs of linear inequalities on a coordinate plane.
Graphical : Graph f(x)≥x−1 .
b.

Answer to Problem 42PPS
Explanation of Solution
Given: Graph f(x)≥x−1 .
Calculation:
f(x)=x−1 has a slope of 1 and a y -intercept of -1 so plot a point at -1 on the y -axis .Then go up 1 right 1 and plot a second point .Continue going up 1right 1 to plot additional points.
Conclusion:
c.
You will investigate the graphs of linear inequalities on a coordinate plane.
Graphical : Plot each point from the table that made f(x)≥x−1 , a true statement on the graph in red .Plot each point that made f(x)≤x−1 , a true statement in blue.
c.

Answer to Problem 42PPS
Explanation of Solution
Given: Plot each point from the table that made f(x)≥x−1 , a true statement on the graph in red .Plot each point that made f(x)≤x−1 , a true statement in blue.
Calculation:
Conclusion:
d.
Logical : Make a conjecture about what the graphs of f(x)≥x−1 , and f(x)≤x−1 , look like. Complete the table with other points to verify your conjecture.
d.

Answer to Problem 42PPS
The graph of f(x)≥x−1 , is the graph of f(x)=x−1 , and then shaded above the line. The graph of f(x)≤x−1 , is the graph of f(x)=x−1 , and then shaded below the line .
Explanation of Solution
Given: Make a conjecture about what the graphs of f(x)≥x−1 , and f(x)≤x−1 , look like. Complete the table with other points to verify your conjecture.
Calculation:
The graph of f(x)≥x−1 , is the graph of f(x)=x−1 , and then shaded above the line. The graph of f(x)≤x−1 , is the graph of f(x)=x−1 , and then shaded below the line .Complete the table with other to verify that this is true:
Point | f(x)≥x−1 . | True/false | f(x)≤x−1 . | True/false |
(5,0) | 0≥5−1 | false | 0≤5−1 | true |
(5,2) | 2≥5−1 | false | 2≤5−1 | true |
(0.1) | 1≥−0−1 | true | 1≤0−1 | false |
(-2,-1) | −1≥−2−1 | true | −1≤−2−1 | false |
(1,-2) | −2≥1−1 | false | −2≤1−1 | true |
e.
Logical : Use what you discovered to describe the graph of a linear inequality.
e.

Answer to Problem 42PPS
If a linear inequality is of the form f(x)≥mx+b, then it will be the graph of f(x)=mx+b, then shaded above the line . If it is in the form f(x)≤mx+b, then it will be the graph of f(x)=mx+b, then shaded below the line.
Explanation of Solution
Given: : Use what you discovered to describe the graph of a linear inequality.
Calculation:
If a linear inequality is of the form f(x)≥mx+b, then it will be the graph of f(x)=mx+b, then shaded above the line . If it is in the form f(x)≤mx+b, then it will be the graph of f(x)=mx+b, then shaded below the line.
Chapter 5 Solutions
Algebra 1
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