Question 4 of the assignment
specifically looks at the polynomial equation

y = *a*x^{2} + 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 x^{2} 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 = *a*x^{2} + x +
2. By taking the
first derivative of this equation we get y′ = 2*a*x
+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.