EMAT 6680 - Exploration 02-2 Write-up - Varying Quadratic Coefficients

David Hornbeck

September 9, 2013

We will here first examine quadratic functions of the form

where a,b are fixed and c [-4,4].
We will consider g(x) = ax2 + bx + c as a function resulting from a transformation of f(x) = ax2 + bx. Any such function f(x) = ax2 + bx is a parabola with roots

          √ --
     - b -  b2    - b
r1 = ---------=  ---&
        2a       a



We can calculate the axis of symmetry by averaging the values of the roots:

1(r1 + r2) = x = (1)--b = --b
2                2  a    2a

Now, let’s consideration the function g(x) = ax2 + bx + c. We know that the mapping f(x)↦→f(x) + c is a vertical translation of f(x) by c units (upward if c > 0, downward if c < 0).
Intuitively, f(x) and g(x) should have the same axis of symmetry. We can check as follows. The roots of g(x) are:

          √ --------
     - b+   b2 - 4ac
r1 = ------2a-------

          √ --------
     --b----b2 --4ac
r2 =       2a

Averaging the two to get the axis of symmetry,

1-(r1 + r2) = x = (1-)- b---b= - b
2                2   2a      2a

Indeed, f and g have the same axis of symmetry, x = -b
Therefore, f and g are parabolas with the same shape, but g is a vertical shift of f by c units.
The following is an animation of this translation as c ranges within [-4,4].

Now, let us also consider the function

h(x) = g(x - c)

In this form, it is apparent that h is horizontal translation of f by c units (if c > 0, the translation is horizontally in the positive direction; c < 0 and the translation is in the negative direction).
Let us now expand h(x).

h(x) = g(x- c) = a(x2 - 2xc + c2)+ (bx - bc) + c

= (ax2 + bx+  c) + (ac2 - 2axc - bc)

We could have expected that h would have a more complex constant term, as well as a different factor of x (as compared to g(x)).
How does h move as c changes within [-4,4], though? As it is both a vertical translation of c and a horizontal translation of c of g(x), we know that the function - in particular its vertex - should travel along a line with slope c
c = -1.

In investigating this function, it was discovered that the particular line along which h(x) moves as c increases and decreases is quite interesting, as it depends on a,b as opposed to c. We know that the line y = mx + b0 which the translated function moves along as c ranges must have a slope of m = 1. We can then rewrite the equation of this line as -x + y = b0.
We now need a point that any such line will travel through in order to determine its y-intercept.. Let us look at the vertex.
The axis of symmetry for h(x) will lie at the x-value that is the average of the values of the two roots. Using the quadratic formula on

h(x) = (ax2 + bx+ c) + (ac2 - 2axc - bc) = ax2 + (b- 2ac)x + (ac2 + (1- b)c)

we get that

1-          --(b---2ac)
2(r1 + r2) =    2a

The axis of symmetry of h(x) is thus x = -(b-2ac)
--2a--- and the vertex is

   - (b- 2ac )  (b - 2ac)2   - (b- 2ac)2   4a2c2 + 4ac - 4abc
h (---2a-----) =----4a----+ -----2a---- + -------4a--------

   - (b2 - 4acb + 4a2c2)+ 4a2c2 + 4ac- 4abc
=  ---------------------------------------

=  - (b---4ac)

Therefore, the vertex of h(x) is (--(b-2ac)
   4a), and we can solve for the y-intercept of our line by calculating

- x + y = b0

b0 = y - x

        - (b2 --4ac-) --(b--2ac)
⇒  b0 =     4a     -     2a

       2b - b2
⇒ b0 = --4a---

Therefore, given any quadratic g(x) = ax2 + bx + c with varying c, one can determine the line along which the horizontal translation h(x) = g(x-c) solely from the coefficients. In particular, this line is

y = x +   4a

For an animation of the movement of h(x) along this line as c ranges from -4 to 4, see the following: (note: to pause or resume the video, click or double-click, respectively)

WU2 Untitled-3 Writeup2