Question 376780
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\ +\ 2)^2\ =\ 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 2\ =\ \pm4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -6]


This is a circle, so it doesn't have a vertex.  The center is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3, -2)]


And the radius is 5.


Find the *[tex \Large x] intercepts the same way you found the *[tex \Large y] intercepts.  Substitute 0 for *[tex \Large y] and solve.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)^2\ +\ (0\ +\ 2)^2\ =\ 25]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ 3)^2\ =\ 21]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ 3\ =\ \pm\sqrt{21}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3\ +\ \sqrt{21}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 3\ -\ \sqrt{21}]


Roughly 7.6 and -1.6


The *[tex \Large y] intercept points that you have are correct.


The *[tex \Large x] intercept points are


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3\ +\ \sqrt{21},0)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (3\ -\ \sqrt{21},0)]



{{{drawing(
500, 500, -10, 10, -10, 10,
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John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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