Question 331602
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For any quadratic function with real coefficients of the form:


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


The vertex is located at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x_v,y_v\right)]


Where:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_v\ =\ p(x_v)\ =\ p\left(\frac{-b}{2a}\right)\ =\ a\left(\frac{-b}{2a}\right)^2\ +\ b\left(\frac{-b}{2a}\right) +\ c]


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


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


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


and


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


If the calculation under the radical, namely *[tex \Large b^2\ -\ 4ac], is positive, then there are two *[tex \Large x]-intercepts.  If *[tex \Large b^2\ -\ 4ac\ =\ 0], then the *[tex \Large x]-axis will be tangent at the vertex, i.e. the single *[tex \Large x]-intercept will be *[tex \Large \left(x_v,y_v\right)] (see vertex discussion above).  If *[tex \Large b^2\ -\ 4ac\ <\ 0], then the graph does not intersect the *[tex \Large x]-axis anywhere.


Because of symmetry, you can plot one additional point.  At the horizontal distance from the *[tex \Large y]-axis to the vertex on the OTHER side of the vertex, there will be a function value equal to the *[tex \Large y]-coordinate of the *[tex \Large y]-intercept.  That is to say, *[tex \Large (2x_v,c)] is a point on the graph.


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 1\ \ ], *[tex \LARGE b\ =\ 4\ \ ], and *[tex \LARGE c\ =\ 3]


Graph:  Plot the five points discussed above and draw a smooth parabolic curve through them.


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