# Thread: Finding a couple of intercepts and stuff

1. ## Finding a couple of intercepts and stuff

The first is 2x-1=y

Simple enough: y=0, and say x=1. Then two minus one equals 1 for (1,0)

Now x=0 so it's -1 divided by y for negative 1 for (0,-1).

These create a straight line that graphs perfectly.

1. I am doing this right, right?

2. My second issue here is that I can't find a checkpoint for the above to save my life. Nothing makes sense... which in turn leads me to believe I'm not doing this right.

Oh, also: Graph 3y=-5

I understand y=3 or x=1, but I don't get how and why a horizontal line below the x-axis is formed from this. There's no explanation in the book and it wasn't covered in class today. Homework for this section gets turned in right before the test for this section on Tuesday when we come back from the weekend.

Again, thank you for your help. It is greatly appreciated.

2. Originally Posted by Ingersoll
The first is 2x-1=y

Simple enough: y=0, and say x=1. Then two minus one equals 1 for (1,0) Mr F says: When y = 0, x = 1/2. (2x - 1 = 0 => 2x = 1 => x = 1/2).

Now x=0 so it's -1 divided by y for negative 1 for (0,-1).

These create a straight line that graphs perfectly.

1. I am doing this right, right?

2. My second issue here is that I can't find a checkpoint for the above to save my life. Nothing makes sense... which in turn leads me to believe I'm not doing this right.