Fitting an inverse quadratic curve

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SUMMARY

The discussion focuses on deriving an inverse quadratic curve defined by the function f(x) = a / (b*x)^2, which passes through two specified points in the +x,+y quadrant. The key relationship established is that for points P_1 = (x_1,y_1) and P_2 = (x_2,y_2) to lie on the curve, the condition y_1 = k / x_1^2 and y_2 = k / x_2^2 must hold, leading to the equation x_1^2y_1 = x_2^2y_2. By setting b=1, the function simplifies to f(x) = k / x^2, where k is determined by the coordinates of the points.

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  • Understanding of quadratic functions and their properties
  • Familiarity with algebraic manipulation and equations
  • Basic knowledge of coordinate geometry
  • Concept of inverse functions
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  • Study the properties of inverse quadratic functions
  • Learn about curve fitting techniques in data analysis
  • Explore the implications of variable transformations in mathematical functions
  • Investigate the use of constants in function definitions and their impact on graph behavior
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Mathematicians, data analysts, and students studying algebra and geometry who are interested in curve fitting and function analysis.

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If I know 2 points in the +x,+y quadrant, and I want to find the

Code:
f(x) =    a
         --------
         (b*x)^2

curve that passes through both points (a and b are constants).

This is probably either really simple or impossible. :rolleyes:
 
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Yes... this is either really simple or impossible.

Let [tex]k=\frac{a}{b^2}[/tex]. Since [tex]f(x) = \frac{a}{(bx)^2} = \frac{a}{b^2x^2}=\frac{\frac{a}{b^2}}{x^2}[/tex], we can say [tex]f(x) = \frac{k}{x^2}[/tex].

If the two points are given by [tex]P_1 = (x_1,y_1)[/tex] and [tex]P_2 = (x_2,y_2)[/tex], then they both lie on such a curve if and only if [tex]y_1 = \frac{k}{x_1^2}[/tex] and [tex]y_2 = \frac{k}{x_2^2}[/tex]. That is, if [tex]x_1^2y_1=x_2^2y_2[/tex]. If this is the case, then let [tex]k=x_1^2y_1[/tex], and translate to a and b. Since we've got two variables and one equation, we're free to choose one of them. For ease, let's make b=1. Then a=k. So, [tex]f(x)=\frac{k}{x^2}[/tex].
 
Though I can't think how I missed that but I do feel quite stupid now
 

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