Question 1163931
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If the parabola has *[tex \Large x]-intercepts of *[tex \Large (0,0)] and *[tex \Large (1,0)], then it must have factors of *[tex \Large (x\,-\,1)] and *[tex \Large x], so an equation for all parabolas that have these intercepts is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax(x\,-\,1)\ =\ a(x^2\,-\,x)]


For the particular parabola with those intercepts that also includes the point *[tex \Large (2,-2)], it must be the case that *[tex \Large \rho(2)\ =\ -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a((2)^2\ -\ 2)\ =\ -2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -1]


Hence the desired function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ -x^2\ +\ x]


 *[illustration Parabola2intercepts1point.jpg]

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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