hi,

i need to work out the co ordinates of two points that the graph of y=2x-1 passes through.

i don't really know how to do this so could somone explain what i have to do please?!?

thanks!

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- Apr 7th 2011, 10:15 AMandyboy179work out the co ordinates
hi,

i need to work out the co ordinates of two points that the graph of y=2x-1 passes through.

i don't really know how to do this so could somone explain what i have to do please?!?

thanks! - Apr 7th 2011, 10:55 AMe^(i*pi)
Pick any old values of x and put the corresponding y value in.

For example (0,-1) is a point on the line - Apr 7th 2011, 10:59 AMandyboy179
oh so would (1, -1) be one aswell?

- Apr 7th 2011, 11:10 AMTheChaz
- Apr 7th 2011, 11:11 AMe^(i*pi)
No, what makes you say that?

The equation is a straight line. If you want to use x=1 then it would be $\displaystyle (1, [2 \cdot 1 - 1])$ - Apr 7th 2011, 11:15 AMandyboy179
would (-1,0) work?

- Apr 7th 2011, 11:17 AMe^(i*pi)
- Apr 7th 2011, 11:18 AMTheChaz
When x = -1,

y = 2x - 1 becomes

y = 2(-1) - 1 =

-2 -1 = -3

S0 (-1, -3) is the point - Apr 7th 2011, 11:19 AMandyboy179
oh i see now, thanks alot!

- Apr 7th 2011, 11:32 AMandyboy179
no its -1

- Apr 7th 2011, 11:42 AMveileen
Uhm, you don't simply search numbers that verify the equation, but take the intersection with the axes (make x=0 and find y, then make y=0 and find x).

- Apr 7th 2011, 11:47 AMNOX Andrew
That's only necessary to find the x- and y-intercepts. For any two points on the line, any two values of x suffice.

- Apr 7th 2011, 10:44 PMCaptainBlack
- Apr 7th 2011, 10:45 PMCaptainBlack
- Apr 7th 2011, 10:45 PMCaptainBlack