Question 1122315
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The standard form of the equation of a parabola with at vertical axis of symmetry is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax^2\ +\ bx\ +\ c]


Since the *[tex \Large y]-intercept of this parabola is at the point *[tex \Large (0,4)], we know that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a(0)^2\ +\ b(0)\ +\ c\ =\ 4]


in other words, *[tex \Large c\ =\ 4]


Also, since the *[tex \Large x]-coordinate of the vertex of a parabola is given by *[tex \Large -\frac{b}{2a}], and the *[tex \Large x]-coordinate of this particular vertex is *[tex \Large 0], we know that *[tex \Large b\ =\ 0]


Now we are left with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ ax^2\ +\ 4]


We also know that the roots of this equation are *[tex \Large -\frac{5}{2}] and *[tex \Large \frac{5}{2}].


Substituting either root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\(\frac{5}{2}\)^2\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\(\frac{25}{4}\)\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -\frac{16}{25}]


Putting it all together is one sock:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ -\frac{16}{25}x^2\ +\ 4]
								
								
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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