Question 626175
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The vertex of the parabola defined by *[tex \LARGE \rho(x)\ =\ ax^2\ +\ bx\ +\ c] is the point *[tex \LARGE \left(x_v,y_v\right)] where *[tex \LARGE x_v\ =\ \frac{-b}{2a}] and *[tex \LARGE y_v\ =\ \rho(x_v)].


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


The maximum or minimum of *[tex \LARGE \rho(x)] is *[tex \LARGE y_v].  If the lead coefficient is positive, i.e. *[tex \LARGE a\ >\ 0], then the parabola opens upward and the vertex is a minimum.  If the lead coefficient is negative, i.e. *[tex \LARGE a\ <\ 0], the parabola opens downward and the vertex is a maximum.


Graph:  Plot the vertex.  Plot the point *[tex \LARGE (0,c)] which is the *[tex \LARGE y]-intercept.  Then by symmetry, *[tex \LARGE (2x_v,c)] is also on the graph.  Set the function equal to zero and solve for the two roots *[tex \LARGE \alpha_1] and *[tex \LARGE \alpha_2] (use the quadratic formula or complete the square since your example doesn't factor).  Plot the two points, *[tex \LARGE \left(\alpha_1,0\right)] and *[tex \LARGE \left(\alpha_2,0\right)].  That gives you 5 points through which you can draw a smooth curve.


If you want more points, pick a convenient distance *[tex \LARGE \delta] from the vertex.  Then calculate the value of the function *[tex \LARGE \rho(x_v\ +\ \delta)], then plot the point *[tex \LARGE \left(x_v\ +\ \delta,\rho(x_v\ +\ \delta)\right)].  Then, by symmetry you can also plot *[tex \LARGE \left(x_v\ -\ \delta,\rho(x_v\ +\ \delta)\right)].  You can repeat this process with as many different values of *[tex \LARGE \delta] as you like.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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