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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Pick any old values of x and put the corresponding y value in. For example (0,-1) is a point on the line
oh so would (1, -1) be one aswell?
Originally Posted by andyboy179 oh so would (1, -1) be one aswell? No. (x, y) = (1, -1) is NOT on the line since using these values in y = 2x - 1 gives us -1 = 2(1) -1, which is FALSE.
No, what makes you say that? The equation is a straight line. If you want to use x=1 then it would be $\displaystyle (1, [2 \cdot 1 - 1])$
would (-1,0) work?
Originally Posted by andyboy179 would (-1,0) work? Sub it into the equation and check. Does $\displaystyle (2 \times -1) -1$ equal 0?
When x = -1, y = 2x - 1 becomes y = 2(-1) - 1 = -2 -1 = -3 S0 (-1, -3) is the point
oh i see now, thanks alot!
no its -1
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).
That's only necessary to find the x- and y-intercepts. For any two points on the line, any two values of x suffice.
Originally Posted by andyboy179 oh so would (1, -1) be one aswell? Quote what you are responding to otherwise the thread will become confusing. CB
Originally Posted by e^(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 $\displaystyle (1, [2 \cdot 1 - 1])$ Quote what you are responding to, otherwise the thread will become incomprehensible. CB
Originally Posted by andyboy179 would (-1,0) work? Quote, or don't bother replying thank you. CB
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