Question 4 of the assignment specifically looks at the polynomial equation

                                      y = ax² + x + 2

where a is varied and b = 1 and c =2.  The graph below shows five different graphs where a = -2, -1, 0, 1, 2.

By looking at the graphs it is obvious that they all share a common point (0, 2) which is the y-intercept for each graph.  The graphs are all parabolas, except when a = 0.  In this case there is no x² term, so this equation is a linear equation, y = x + 2. 

The equation y = x +2 is also the tangent line at the point (0, 2) of every equation in the form y = ax² + x + 2.  By taking the first derivative of this equation we get y′ = 2ax +1 and substituting x = 0 we have the slope of the tangent line is equal to one.  Since the point is (0, 2) we get the equation y = 1(x 0) + 2, or

                     y = x + 2.

Since the vertex of each parabola changes, there is a vertical and horizontal shift with each new a value that is used. 

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