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Explain this to me PLEASE! (algebro)?
Watch this video please about 30 seconds in or so there is no timer so you'll have to watch it until you get to the part where he's doing step 1/2 (graphing)
http://www.phschool.com/atschool/academy123/englis...
I AM SOOO Confused. If the problem is set up like this:
y =2x - 1
y = -x + 5
Why on earth is it that for the first problem he graphs like
^2(x) >1(y)
But for the second problem he goes
^5(y) v> x (x)
Its confusing me to no end!!! Help me please I have an exam soon and I need to know this.
I'm trying to find out why is it that the first problem he moves up the x value and then moves sideways with the y value then does vice versa for the equation beneath. -Moves up the y value then sideways the x
1 Answer
- SamwiseLv 77 years agoFavorite Answer
The equations both are in the general form we call "slope-intercept form":
y = mx + b
where m is the slope of the line
and b is the y-intercept.
For y = 2x - 1, then, the slope is 2 and the y-intercept (the point on the y-axis where the line crosses it) is at (0,-1). So he marks that point and then starts marking points that would be on a line, going through it, with slope 2.
He does this by moving up 2 units and over 1 unit each time (that is, adding 2 in the y-dimension for every 1 in the x-dimension). From our starting point, this gives a sequence of points
(0,-1), (1,1), (2,3), (3,5) and so on. These points are all on the line with this equation, so he draws the line through them.
For equation y = -x + 5, we have slope -1 and y-intercept 5: it crosses the y-axis at (0,5). Starting there, he uses the slope to find additional points. Since the slope is -1, each addition of 1 to x means subtracting 1 from y, so the points are in the sequence:
(0,5), (1,4), (2,3), (3,2), (4,1) and so on. The second equation defines a line that goes through these points.
From your last sentence, I'm wondering whether your confusion arose from misunderstanding how he was using x and y. In both cases, his procedure for each new point was:
(step 1) add 1 to x;
(step 2) add the value of the slope to y.
That works because slope is defined as
(difference in y values)/(difference in x values),
so if the difference in x values is always 1, the difference in y values is always the number we have for the slope.
