Question 511129
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The domain of any polynomial function is the set of real numbers.  Zero is a real number, therefore for any polynomial function *[tex \Large f], *[tex \Large f(0)] is defined and exists.  The *[tex \Large y]-intercept of a polynomial function *[tex \Large f] is the point *[tex \Large \left(0,f(0)\right)].


*[tex \Large x]-intercepts of the graph of a polynomial function are points of the form *[tex \Large (x_i,0)] where *[tex \Large x_i] is a real number zero of the polynomial function.  Since complex roots always occur in conjugate pairs, that is to say if *[tex \Large a\ +\ bi] is a root, then *[tex \Large a\ -\ bi] must also be a root, and the number of zeros of a polynomial function is equal to the degree of the function (Fundamental Theorem of Algebra) it is possible to have a polynomial function of even degree with no real number zeros, and therefore no *[tex \Large x]-intercepts.  Polynomial functions of odd degree are guaranteed to have at least one real zero and therefore at least one *[tex \Large x]-intercept.


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