# Thread: Graphing inequalities minor problem..

1. ## Graphing inequalities minor problem..

I have completed every problem except this one, and the only problem i'm having is the start.. if i get it off the ground, i can probably finish it.
We are graphing inequalities, (y'know, draw the line, pick a point, prove/disprove, shade true area), and i have come to one that doesnt seem to have a line point. all the others had a +1 or -3 or +4 or something like that, but this one is just

y < 2x

am i supposed to just put in +0 for a starting point, or is there something more?
i have to go to bed now, so don't expect responses until morning, but i will log on to check and see what to do..it is, after all, due tomorrow in school! :P

2. Originally Posted by Nightfire
I have completed every problem except this one, and the only problem i'm having is the start.. if i get it off the ground, i can probably finish it.
We are graphing inequalities, (y'know, draw the line, pick a point, prove/disprove, shade true area), and i have come to one that doesnt seem to have a line point. all the others had a +1 or -3 or +4 or something like that, but this one is just

y < 2x

am i supposed to just put in +0 for a starting point, or is there something more?
i have to go to bed now, so don't expect responses until morning, but i will log on to check and see what to do..it is, after all, due tomorrow in school! :P
To draw the line y = 2x you need two points. Then you join them with a line. And yes indeed, one of those points is (0, 0).

To get another point, sub x = 1 into y = 2x: y = 2(1) = 2. So another point is (1, 2).

Plot both points and draw the line. Now decide which side of the line your inequality satisfies .....

Oh, how did I choose x = 1 to get the second point I hear you ask. I could've chosen any value of x. But x = 1 is real simple .....

3. Originally Posted by Nightfire
I have completed every problem except this one, and the only problem i'm having is the start.. if i get it off the ground, i can probably finish it.
We are graphing inequalities, (y'know, draw the line, pick a point, prove/disprove, shade true area), and i have come to one that doesnt seem to have a line point. all the others had a +1 or -3 or +4 or something like that, but this one is just

y < 2x

am i supposed to just put in +0 for a starting point, or is there something more?
i have to go to bed now, so don't expect responses until morning, but i will log on to check and see what to do..it is, after all, due tomorrow in school! :P
If I understand your question correctly, you were expecting to see an inequality that, if written with an equals sign, satisfies the form y=mx+b where m is the slope and b is the y-intercept. Since y<2x is void of a "b" you weren't sure if it should be graphed differently than the others. I'm sure you have plenty of experience graphing lines and using different methods to graph them, so just think of these inequalities as lines - the only real difference between them and lines is that the inequality forces you to shade the graph, but all the other standards for graphing lines still apply.

4. @ecMathGeek: no, not on these. all these are done in the style of y > 1x + 1 or something along those lines, no B involved. i'm not using a method involving slope-intercept form.
@mr fantastic: alright, thats what i thought it was. thanks for killing my doubts.