Question 173836
Your problem is that you are beginning with a quadratic equation and confusing that with a function whose graph is in the shape of a parabola.  Furthermore, you have made a sign error when you rearranged the equation by putting everything on the right (although I suspect this was simply a typo when you posted this problem, otherwise you would not have obtained the same pair of roots as your instructor did).  {{{-2x^2=16x+28}}} is equivalent to {{{0=2x^2+16x+28}}}.  Furthermore, this is equivalent to  {{{-2x^2-16x-28=0}}} because it is perfectly valid to multiply both sides of an equation by -1 and completely swapping sides is perfectly valid as well.


The simple fact is that all of the above equations have a solution set that consists of the two roots of the equation.  They do NOT represent a parabola which is a curve that consists of an infinite number of points.


Now, if you want to look at the parabola associated with the quadratic equation, you can always write  {{{f(x)=y=2x^2+16x+28}}} which is, in fact a parabola whose vertex is a minimum.  NOW if you multiply by a -1, you get: {{{-f(x)=-y=-2x^2-16x-28}}} and, as you might expect, your graph will be a parabola with the vertex as a maximum.  The two functions do, in fact, have the same x-intercepts as you would expect from the results of solving the original quadratic equation.



{{{graph(600,600,-10,10,-10,10,2x^2+16x+28,-2x^2-16x-28)}}}