Question 1188454
Sunday, April 2, 2017
7:06 PM

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Quadratic with roots 3 and -5 that passes through (1,5).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ a(x\,-\,3)(x\,+\,5)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(1)\ =\ a(-2)(6)\ =\ 5]


Hence *[tex \Large a\ =\ -\frac{5}{12}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ -\frac{5}{12}x^2\ -\ \frac{5}{6}x\ +\ \frac{25}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d\rho}{dx}\ =\ -\frac{5}{6}x\ -\ \frac{5}{6}]


Set the first derivative equal to -4 and solve:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{5}{6}x\ -\ \frac{5}{6}\ =\ -4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 1\ =\ -4\(-\frac{6}{5}\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{19}{5}]


All lines with slope -4 have an equation of the form *[tex \Large y\ =\ -4(x\,-\,x_1)\ +\ \rho(x_1)].  Since a line with a slope of -4 is tangent to the graph of the quadratic function at the point *[tex \Large \(\frac{19}{5},\rho\(\frac{19}{5}\)\)], or *[tex \Large \(\frac{19}{5},-\frac{44}{15}\)], the line *[tex \Large y\ =\ -4\(x\,-\,\frac{19}{5}\)\ -\ \frac{44}{15}] intersects the function graph exactly once.  Any line of the form *[tex \Large y\ =\ -4(x\,-\,x_1)\ +\ \rho(x_1)] where *[tex \Large x_1\ >\ \frac{19}{5}] will not intersect the function graph anywhere. And where *[tex \Large x_1\ <\ \frac{19}{5}], the line will intersect the function graph in two places.


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

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