更新1:
and i forgot to add, and can a proportional relationship problem have to start at 0,0 (origin) ?
help with slope, y intercept and some other things?
2016-01-14 7:08 pm
Im having a hard time understanding slope and y intercept, how do find both of them in a graph and equation? im asian so im new to america, so how does -2-2 means -4? at first i thought it would be 4, how is it negative :/
回答 (2)
2016-01-14 7:17 pm
-2 subtract 2 is -4
-2 times -2 = +4
Y intercept happens when x =0
X intercept happens when y = 0
-2 times -2 = +4
Y intercept happens when x =0
X intercept happens when y = 0
2016-01-14 7:14 pm
-2 -2 = -4. Loot at like (-2) + (-2). Now, -2 x -2 = 4
As for how you find slope and intercept, it depends on what information you are give. If you have the equation y = mx +b, then m is the slope and b in the y intercept.
If you have a line in any other form, just put it in this form.
so:
ay + bx = c becomes
ay = -bx + c
y = (-b/a) x + c/a, where -b/a is the slope and c/a is the y intercept.
If you have two points (x1,y1) (x2,y2), them the slope is (y1-y2)/(x1-x2). Once oyu have the slope (m), take either point and write the equation:
(y-y1)/(x-x1) = m
y-y1 = m(x-x1)
y = mx -m(x1) + y1, so -m(x1) + y1 is the y intercept.
As for how you find slope and intercept, it depends on what information you are give. If you have the equation y = mx +b, then m is the slope and b in the y intercept.
If you have a line in any other form, just put it in this form.
so:
ay + bx = c becomes
ay = -bx + c
y = (-b/a) x + c/a, where -b/a is the slope and c/a is the y intercept.
If you have two points (x1,y1) (x2,y2), them the slope is (y1-y2)/(x1-x2). Once oyu have the slope (m), take either point and write the equation:
(y-y1)/(x-x1) = m
y-y1 = m(x-x1)
y = mx -m(x1) + y1, so -m(x1) + y1 is the y intercept.
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