Question

See below.

Answer 1

Plot the graphs y=-x, y=3+2x, and y=3. Then shade in the y-values accordingly. I.e. if y<-x, shade in the region above the line y=-x. Do the same for the others and you should end up with a region bounded by all three equations.

Answer 2

ok..hmmm....im having some issues with the plotting part though..

Answer 3

Ok I'l try to help. Do you know the graph of y=x? This equation just means that whenever you have some value for x, the corresponding y value will be the exact same. Meaning you will have points on the graph such as (1,1), (6,6), (-4, -4). Does that make sense? Now, when you have y=-x, the corresponding y value will be the negative of the x value. For example, some points on the line y=-x will be (1,-1), (3,-3), (-2, 2). y=-x is the equation of a line with slope -1 and y-intercept 0

Answer 4

hmm oh...yes.

Answer 5

Do you have to use intercepts (x and y) to plot the points on the graph..?

Answer 6

No you don't have to, especially for linear functions, meaning equations which visually are a line. For a linear funcation all you need is two points on the graph, and you just have to connect those and extend it, then you have your line. However, for more complicated quadratic equations (which on a graph are curvy) the x and y intercepts are essential.

Answer 7

Ok thanks for the info on that. I'll try figuring out that problem now.