2.1.2 Linear Tables and Graphs (Algebra 1)
The question is on the image below


Answer:
Step-by-step explanation:
1). Let the equation of a line is,
y = mx + b
Here, m = slope of the line
b = y-intercept
From the graph attached,
Slope = [tex]\frac{\text{Rise}}{\text{Run}}[/tex] = [tex]\frac{1}{4}[/tex]
y-intercept = -2
Equation → y = [tex]\frac{1}{4}x-2[/tex]
2). Slope of the line = [tex]\frac{\text{Rise}}{\text{Run}}=\frac{3}{4}[/tex]
y-intercept 'b' = -3
Equation → y = [tex]\frac{3}{4}x-4[/tex]
3). Slope of the line = [tex]\frac{\text{Rise}}{\text{Run}}=\frac{-5}{0.75}[/tex]
= [tex]-\frac{20}{3}[/tex]
y-intercept = 5
Equation → y = [tex]-\frac{20}{3}+5[/tex]
4). Slope of the line = [tex]\frac{\text{Rise}}{\text{Run}}=\frac{-2}{4}=-\frac{1}{2}[/tex]
y-intercept = 4
Equation → y = [tex]-\frac{1}{2}x+4[/tex]
5). Since, line is passing through origin (0, 0)
y-intercept = 0
Slope = [tex]\frac{\text{Rise}}{\text{Run}}=\frac{1}{1}[/tex]
Equation → y = x
6). Slope of the line = [tex]\frac{\text{Rise}}{\text{Run}}=\frac{1}{4}[/tex]
y-intercept = -4
Equation → y = [tex]\frac{1}{4}x-4[/tex]