Question 520820
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{1}{4}\left(x\ -\ 2\right)^2\left(x\ +\ 2\right)^2]


To determine the *[tex \Large y]-intercept, which is the value of the function when *[tex \Large x\ =\ 0] substitute *[tex \Large 0] for *[tex \Large x] and do the arithmetic:  *[tex \Large \left(0,\,4)]


To find the *[tex \Large x]-intercepts, find the zeros of the function.  Since this is a 4th degree polynomial function, there are 4 zeros. (Fundamental Theorem of Algebra).  The easiest way to see all four zeros is to write out the function in its fully factored form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{1}{4}(x\ -\ 2)(x\ -\ 2)(x\ +\ 2)(x\ +\ 2)]


Clearly the zeros are 2 with a multiplicity of 2 and -2 with a multiplicity of 2, for a total of four zeros counting the multiplicities.


A zero that has a multiplicity of 2 occurs where the graph has the *[tex \Large x]-axis as a tangent.



{{{drawing(
500, 500, -6,6,-6,6, 
grid(1),
graph(
500, 500, -6,6,-6,6, 
(1/4)*((x-2)^2)((x+2)^2) ))}}}


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