Explorations with Some Second Degree Equations

by Asli Ersoz

Problem: Graph

i. Overlay a new graph replacing each x by (x - 4).

ii. Change the equation to move the graph into the second quadrant.

iii. Change the equation to produce a graph concave down that shares the same vertex.

First of all, here is the graph of

:

Here is what happens when I replace each x by (x-4):

So, the graph moves 4 units to the right. The question is; why does the graph behave like this? Let's think of a x-value: when x = 0 y = -4 in the first equation. when is y = -4 in the second equation? It is when x = 4.

So, what should we do to move the graph into the second quadrant? Based on our observation above, we should replace x by (x + b) (b is positive) to move the graph to the left. How many units of replacement do we need?

Try this:

However, this was not enough. We also need to do something to move the graph up. This is something we can do by varying (-4). Keeping the coefficients of and x same and varying the constant term translates the graph horizontally, i.e. moves up or down. But, how much should we change the constant term? Observe that the y-value of the vertex of the graph is below -5. So, increasing -4 by 6 should work:

Our last task is changing the equation so that the new graph will be concave down and share the same vertex with the original one.

Let's do this step by step:
For a graph to be concave down the coefficient of the should be negative. So, let's graph

However, this not only turned the graph upside down but also translated it to the right. To reflect it over the y-axis and 'bring back' replace x by (-x).

Now, we have to move it down. We know that we need to change the constant term (-4) but by how much? We know that the coordinates of the vertex of a parabola can be found in the following way:

Set f'(x) = 0, solve for x and substitute this value for x in f(x).

f(x) = a + bx + c

f'(x) = 2ax + b = 0 x = -b/2a