Question 728790
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For any quadratic of the form *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c], the graph is a parabola with a vertical line axis of symmetry opening upward if *[tex \LARGE a\ >\ 0] or downward if *[tex \LARGE a\ <\ 0].


The *[tex \LARGE x]-coordinate of the vertex is given by *[tex \LARGE x_v\ =\ \frac{-b}{2a}].  The *[tex \LARGE y]-coordinate of the vertex is *[tex \LARGE \rho(x_v)\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c]


The axis of symmetry is the line with the equation *[tex \LARGE x\ =\ x_v]


The *[tex \LARGE y]-intercept is the point *[tex \LARGE (0,\,c)]


Symmetry demands that there is a point on the graph at *[tex \LARGE (2x_v,c)]


The *[tex \LARGE x]-intercepts, if they exist (and they do exist for your problem) are at *[tex \LARGE (\alpha_1,\,0)] and *[tex \LARGE (\alpha_2,\,0)] where *[tex \LARGE \alpha_1] and *[tex \LARGE \alpha_2] are the zeros of the polynomial.  In other words, *[tex \LARGE \rho(\alpha_1)\ =\ 0] and *[tex \LARGE \rho(\alpha_2)\ =\ 0].  Note that your trinomial factors neatly over the integers:  -3 + 2 = -1 and -3 X 2 = -6.


Plot the above five points and draw a smooth curve.


For future reference, any quadratic with a non-zero value for the constant term and the constant and lead coefficient have opposite signs, you are guaranteed two distinct real number zeros.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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