Graphing Linear Equations: Table of Values, Slope, and Equations

This assignment focuses on graphing linear equations using two primary methods: the table of values and the slope-intercept form. The solution begins by demonstrating how to construct a table of values to plot points and graph linear equations, emphasizing the importance of choosing appropriate x-values to accurately represent the line. Several examples are provided, each with a corresponding table and a graphical representation. The second part of the assignment delves into graphing using the slope and y-intercept method, providing detailed steps and examples to illustrate the process. The solution also includes an example of how to represent a real-world scenario, such as the cost of a lawn care service, with a linear equation and its corresponding graph. Overall, the assignment offers a comprehensive guide to graphing linear equations, suitable for students studying algebra.

Activity 8: Graphing by hand- table of values
Discussion
how important is choosing correct x-values to make a table of values
Choosing correct x-values in tabulating values of linear equation is important as it predertermine
what the shape and position of the line values on the cartesian plane.
When creating a graph, it is important for a drawn line graph to cover all the sections of the
cartesian and or the cartesian. This is because when it covers all edges, it will be easy to
determine and point out both x and y intercepts
Part A
Solution
a) Y = 2x – 1
b) Y = -3x +4

c) Y = ½ x +1
d) x + y =2

-3 -2 -1 0 1 2 3 4 5
-8
-6
-4
-2
0
2
4
6
f(x) = − 2 x + 1
e) 3x - y = 9
Activity 9: Graphing by hand – slope and y-intercept
Solution
Part A
Graphing the equations on paper

a) Y = 2x
Construct a table as shown
x y
-2 -4
0 0
2 4
4 8
Then the graph is plotted as shown
b) Y = -2x +3
Construct a table as shown
x y
-2 7
0 3
2 -1
4 -5

Then the graph is plotted as shown
c) Y = 1/3x – 1
The table is constructed as shown
x y
-2 -1.66667
0 -1
2 -0.33333
4 0.333333
The graph is drawn as shown

-3 -2 -1 0 1 2 3 4 5
-8
-6
-4
-2
0
2
4
6
f(x) = − 2 x + 1
d) Y = -3/2 x +2
The table for the equation is
x y
-2 5
0 2
2 -1
4 -4
The graph is drawn as shown

e) 2x +y =1
The table becomes,
x y
-2 5
0 1
2 -3
4 -7
The graph is drawn as
Part B
1. Equation to represent the cost to hire
Let the cost to hire be represented by, C
Time to be represented by, t
The equation become: C=20 t+ 90

2. Graph for the relationship
3. How much the lawn care service cost
equation of cost ,C=20t +90
time required per week=1.5 hours
care service cost per week=20∗1.5+90=$ 120
care service cost for 6 weeks=$ 120∗6=$ 720
care service cost for 4 weeks=$ 120∗4=$ 480
steps needed to graph a linear equation using a slope and y-intercept method
Taking an example of the linear equation to be y= 1
4 x−3

The first step is to write the equation In the form of y = mx + c