Looking for an equation expressed algebriacally that answers the following

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The discussion revolves around finding an algebraic equation that expresses the value of B in relation to A on a Cartesian plane, specifically when A is defined as a horizontal line at y=1. The user seeks a formula for B that starts at the origin (0,0) and approaches A at a constant rate. A participant clarifies that the interpretation of "approaches infinite A" is likely a misunderstanding, suggesting that B can be modeled using the equation y=1+Ce^{\alpha x}, where α is a negative constant, indicating that B asymptotically approaches y=1 as x increases.

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seasnake
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Given an XY graph where the horizontal line is X and the vertical line is Y and their intersection is zero, if A = 0,1 (a horizontal line one increment above the X axis), I want to know the formula that correctly expresses the value of B if B started at 0,0 and always approaches infinite A at a uniformly constant rate.
 
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Given an XY graph where the horizontal line is X and the vertical line is Y and their intersection is zero
a.k.a. The Cartesian Plane. Probably the most widely known and frequently used coordinate system in Mathematics :smile:

Everything is very clear cut, until...
approaches infinite A

eh? Please elaborate.
 
seasnake said:
Given an XY graph where the horizontal line is X and the vertical line is Y and their intersection is zero, if A = 0,1 (a horizontal line one increment above the X axis), I want to know the formula that correctly expresses the value of B if B started at 0,0 and always approaches infinite A at a uniformly constant rate.
You just said "A= 0, 1", a notation I would have interpreted as the point (0, 1) but then you say "(a horizontal line one increment above the X axis)" which is the line y= 1. In either case, what do you mean by "approaches infinite A"? A is NOT "infinite". Do you mean "is asymptotic to y= 1 as x goes to infinity"? And what do you mean by "approaches at a uniformly constant rate"? That d(y-1)/dx= constant? That's impossible. Any solution to that is linear and cannot be asymptotic to y= 1. d(y-1)/dx= constant*(y-1) is possible. Calling the constant rate [itex]\alpha[/itex], any solution to that is of the form
[tex]y=1+ Ce^{\alpha x}[/tex]
which approaches y= 1 at constant rate [itex]\alpha< 0[/itex].
 

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