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May 24th, 2003, 02:59 PM
#1
Thread Starter
Dazed Member
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May 25th, 2003, 05:36 AM
#2
Addicted Member
hey mate,
your equation is the equation of a straight line....
y = mx + c where m = gradient, c = y intercept
it is derived from the definition of the gradient;
m = (change in y) / (change in x) = (y-y1)/(x-x1)
therefore m(x-x1) = y-y1
y = m*x - m*x1 + y1 = m*x + (y1- m*x1)
when x = 0, y = y1-m*x1, this is how we know it is the y intercept
if y1 and x1 are known then the whole thing can be reduced to the constant c. when m is negative, the function is a decresing one and when it is positive the function is an increasing one.
feel free to ask anything if you still don't understand.
One thing that sustains me through life is the conciousness of the immense inferiority of everyone else
--Oscar Wilde
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May 25th, 2003, 07:37 PM
#3
Thread Starter
Dazed Member
I figured it out. A direct relationship uses dependentvar = ver intercept + slope * independent var and an inverse relationship uses independentvar = verintercept + slope * dependentvar.
Since i am dealing with graphs there is no need to calculate the vertical intercept. Since the vertical intercept of a line is the place where it meets its vertical axis all i have to do is look at the graph.
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May 25th, 2003, 09:04 PM
#4
Fanatic Member
if I understand your question you want to find where two lines intersect.
as said above: Y = MX + B
The trick to finding it is making the equations = to each other. (both of them = to Y or 2Y etc...)
So, if, for example, they gave you:
2Y = 6X + 4
Y = .4X - 15
First, isolate Y (already done) and make it = to the Y in other equation:
2Y/2 = 6x/2 + 4/2
Y = 3x+2 and y= .4x - 15
No, now we know that: 3x+2 = .4x - 15
Isolate X:
3x-.4x + 2-2 = .4x/.4x-15-2
or
2.6x = -17
17/2.6 = ~6.53846...
So we know that the X where they intercept is 6.5, now we can find Y from either of the original equations
2Y = 6X + 4
2Y / 2 = ( 6(6.5) + 4) / 2
Y = 21.5
So they intercept at (6.5, 21.5)
Don't pay attention to this signature, it's contradictory.
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