Question 1183273
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The two zeros can be found by setting each of the factors to zero and then solving for *[tex \Large x].  Example:  If one of the factors was *[tex \Large x\ -\ a], then the associated zero would be found by setting *[tex \Large x\ -\ a\ =\ 0] with the resulting zero being *[tex \Large x\ =\ a].


Once you have found the two zeros of a quadratic, the *[tex \Large x]-coordinate of the vertex is exactly halfway between the two.  For example, if the two zeros were *[tex \Large x\ =\ a] and *[tex \Large x\ =\ b], then the *[tex \Large x]-coordinate of the vertex would be *[tex \Large x_v\ =\ \frac{a\,+\,b}{2}].


The *[tex \Large y]-coordinate of the vertex is simply the value of the function at the *[tex \Large x]-coordinate you calculated as described above.  Just substitute that *[tex \Large x_v] value into the original equation (either the factored form or the unfactored form, you get the same answer either way given correct arithmetic) and do the indicated arithmetic.

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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