Sarah Hofmann EMAT 6680 Summer 2006 Assignment 3 Investigation 1 Our assignment for investigation 1 was to notice that all equations of the form have their verticies on the equation as shown in the graph below. We now need to generalize this idea. Investigation 2 We were to graph the equation where a = 1 in the xb plane for different values of c.  We have chosen to plot the integers c = -4 . . . 3.  Notice that in the case where c = 0, the turquoise line, is linear. Investigation 3 Our job for this investigation is to add the equation to the above graph and investigate its relationship to the quadratic formula.  We will add the equation to the above graph in the form .  It will appear as a purple line.   The quadratic formula for our equation with a = 1 is . If we look at the quadratic equation in the context of the line , we can see the following But we know that implies that the equation has only one root. Investigation 4. We are to consider graphs in the xc plane for this investigation. If we graph several basic parabolas in the form  for integer values of c=-4...3, we can see the following graph. If we differentiate the basic equation, solve for y', and set y' = 0, we can see the first derivative is zero when .  So the local maxima occur on this parabola.  Indeed if we add the equation to the above graph we see it intersects all the parabolas at their verticies.  The new parabola is in purple. Now if we consider graphs of the form for values of a = -4...3 we can see the following graphs. Differentiating the equation , solving for y', and setting y'=0 we see that the critical points all coincide on the line .  Indeed if we add this line as a purple line and zoom in to the above graph we can see it does cross all the local maxima or minima of the above parabolas. Investigation 5. We are to consider graphs in the xa plane. If we graph several graphs of the basic equation for values of c=-4...3 we get the following graphs. If we consider the basic equation for integer values of b=-4...3, we get the following graphs. Differentiating the basic equation , solving for y', setting y'=0, and solving, we see that all the critical points lie on the equation .  Adding this equation as a purple graph to the above graph we see that indeed, all the critical points pass through the new purple line.