Question 1164128
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I will show you how to do this in general, and then you can do this problem and any one you encounter like it in the future:


Given:


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


The *[tex \Large x]-coordinate of the vertex is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_v\ =\ -\frac{b}{2a}]


The *[tex \Large y]-coordinate of the vertex is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x_v)\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c]


And the equation of the axis of symmetry is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ x_v]


Bonus fact: The *[tex \Large x]-intercepts, if they exist, are located at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(x_v\ \pm\ \frac{\sqrt{b^2\ -\ 4ac}}{2a},0\)]


and the  *[tex \Large y]-intercept is located at *[tex \Large (c,0)]


																
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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