Question 611383
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Rewrite your equation with everything except the y in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -x^2\ -\ 2]


Compare this function to the vertex form of the parabola function:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 4p(x\ -\ h)^2\ +\ k]


which is a parabola with vertex at (h,k), focus at (h,k + p), axis of symmetry at *[tex \LARGE x\ =\ h], and *[tex \LARGE y]-intercept at (0,k).


Since your equation has no 1st degree x term, *[tex \LARGE h\ =\ 0].  By inspection, *[tex \LARGE 4p\ =\ -1] hence *[tex \LARGE p\ =\ -\frac{1}{4}], so let's re-write again:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 4\left(-\frac{1}{4}\right)(x\ -\ 0)^2\ +\ (-2)]


Now we can see that the vertex is at *[tex \LARGE (0,-2)], the focus is at *[tex \LARGE (0,-2.25)], the *[tex \LARGE y]-intercept is the same as the vertex.


Next pick a couple of positive values (small positive integers) and substitute those values for *[tex \LARGE x].  Do the arithmetic to find the function value at those points.  Plot the points to the right of the *[tex \LARGE y] axis. Use the property of symmetry to find the corresponding points on the left side of the *[tex \LARGE y] axis.


Finally draw a smooth curve through all of your points.


Or skip all of that stuff and use a graphing calculator.


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