Question 674435
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Parabolas do not have "solutions".  Parabolas are graphs of quadratic functions, the *[tex \LARGE x]-coordinates of the *[tex \LARGE x]-intercepts of which are the zeros of the function.  If you set the function equal to zero creating a quadratic equation, then the zeros of the function become the solutions of the equation.


If *[tex \LARGE \alpha] is a zero of a polynomial function in *[tex \LARGE x], then *[tex \LARGE x\ -\ \alpha] is a factor of the polynomial.


So for your question a), the factors of your desired function are *[tex \LARGE x\ -\ (-5)] and *[tex \LARGE x\ -\ 8].  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ (x\ +\ 5)(x\ -\ 8)]


You can do the FOIL and simplify process yourself.


Note for part b) *[tex \LARGE f(\alpha)\ =\ 0] is the same thing as saying *[tex \LARGE \alpha] is a zero of the function


Note for part c) Complex zeros of polynomial functions with real number coefficients ALWAYS come in conjugate pairs.  That is to say, if *[tex \LARGE a\ +\ bi] is a zero of a given polynomial, then *[tex \LARGE a\ -\ bi] is also a zero of that polynomial.


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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