I am very confused on this problem http://i170.photobucket.com/albums/u...g?t=1264619788

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- Jan 27th 2010, 11:17 AMpletcheraAlgebra y=mx+b
I am very confused on this problem http://i170.photobucket.com/albums/u...g?t=1264619788

- Jan 27th 2010, 11:28 AMsaumaun
y=mx+b is "Slope-Intercept Form" where "y" is the equation of the line, "m" is the slope of the line, and "b" is the y-intercept.

To find the y-intercept (or "b"), you should look for the point where the line crosses the y-axis. On this graph, the line crosses the y-axis when y=0, so b=0.

To find the slope (or "m"), you can use the trick "rise/run". Basically, find 2 points on the line that you know. You already have (-2,-1) and (2,1). To get from the lower point to the higher point, you must go UP 2 units. To get from there to the point you must go RIGHT 4 units. So you would have a slope of 2/4 which reduces to 1/2.

Therefore, you equation for the line would be y=1/2x - Jan 27th 2010, 11:30 AMbigwave
when you are given 2 points first find the slope

then use*point slope form*

after simplifying then you will have the*slope-intercept form*

Given: (-2,-1),(2,1)

- Jan 27th 2010, 11:45 PMHallsofIvy
Yet another way to do this: Any non-vertical line can be written as y= ax+ b. Since the line passes through (x, y)= (2, 1) we must have 1= a(2)+ b. Since the line passes through (-2, -1), we must have -1= a(-2)+ b. Solve the two equations 2a+ b= 1 and -2a+ b= -1 for a and b.