Implicit/Elliptical Curve
Assignment Number 3 Due for Sept. 29
Graphing f(x) yielded a regular linear graph with a slope of 1.
However when we graphed the second function (y^2)=(x^3)+ax, it yielded an implicit function, meaning that instead of getting your regular input f(x)= some y output, x and y must be treated as inputs yielding a zero as an output. So f(x,y)=0.
-One of the components for the implicit function included 'ax' for which a slider was provided to vary 'a.'
-When 'a' is animated the intersections between f(x) and the implicit curve change, in fact more intersections b, and c occur when 'a' goes to a negative value -0.7. As soon as a<0, a circular shape is formed along the second and third quadrants of the plane. This has to do with the roots of the equations for example, since x^3 was given there's an implication of three solutions.
What is algebaically happening is that there's solutions occuring for f(x) and the implicit curve. there can only be one solution between them when a>0, because there are values inputed that coincide with f(x). When you try to solve for these separate x and y functions you must make an effort to isolate the variables. So if you were to solve for y first you're left with y=[(x)((x^2)+a)]^(1/2). There could be point where there's at least three solutions, when you find the zeroes for this equation.
The implicit curve has a value that does not equal zero, and hence you only have one solution for all 'a' values greater than 0.