Graphing linear equations questions 1-13

1. the graph of 2x + 1 = y for all real numbers will be a line. the graph with the restricted domain will be a line segment meaning it has endpoints (1, 3) and (4, 9).

2. is open ended, meaning you can make A, B, C anything you want as long as what they specify is met.
(a) A = 0
(0)x + By = C
let's make B = 1 and C = 5
y = 5

(b) B = 0
Ax + (0)y = C
let's make A = 1 and C = 5
x = 5

(c) C = 0
Ax + By = 0
let's make both A and B = 1
x + y = 0

3. plot both intercepts (x on x-axis, y on y-axis) then draw a line through the dots.

4-7. linear equations do not have exponents on any of the variables. 1/y is the same as y^(-1) so both 4 and 7 are non-linear.

8-13. you will have to graph these. put them in slope intercept form: y = mx+b if they're not already , plot the y-intercept first (0, b) then use the slope m draw the line by going rise over run.

example 13.
3x + 4y = 12
4y = -3x + 12
y = (-3/4) + 3
plot (0,3)
go down -3 , right 4 make a point.
reverse from (0,3) go up 3 left -4 and make a point. connect with line.