Question 322171
<font face="Garamond" size="+2">


The lead coefficient is positive, so it opens up.


There is no 1st degree term, so the vertex is the *[tex \LARGE y]-axis.  Therefore the value of the *[tex \LARGE x]-coordinate of the vertex is 0 and the value of the function when *[tex \LARGE x] is zero is *[tex \LARGE y(0)\ =\ -2(0)^2\ -\ 3\ =\ -3], hence the vertex is at the point *[tex \LARGE \left(0,\,-3\right)].


The axis of symmetry is the vertical line passing through the vertex.  The equation of any vertical line is *[tex \LARGE x\ =\ \alpha] where *[tex \LARGE \alpha] is the *[tex \LARGE x]-coordinate of any point on the line.  Since we know the vertex is on the line and we have already determined that the *[tex \LARGE x]-coordinate of the vertex is 0...


The parabola opens upward, so the vertex is a minimum.


The domain of any polynomial function with real-valued coefficients is the set of real numbers.


The range of a parabola that opens upward is *[tex \LARGE \beta\ \leq\ y\ \leq \infty]


where *[tex \LARGE \beta] is the *[tex \LARGE y]-coordinate of the vertex.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>