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

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

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

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:

Averaging the two to get the axis of symmetry,

Indeed, f and g have the same axis of symmetry, x = .
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

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).

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 = -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

we get that

The axis of symmetry of h(x) is thus x = and the vertex is

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

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

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)

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