Investigating Second Order Polynomials

 

By: Stephen Bismarck

 

 

 

Produce several ( 5 to 10) graphs of

 

y = (x-d)2 - 2

 

on the same axes using different values for d. Does varying d change the shape of the graph? the position?

 

 

The first thing I did for this problem was pick some values for d. I picked d = -10,-3,0,2,3,10

 

 

 

 

 

 

 

y = (x +10)2 – 2                                  y = (x + 3)2 – 2

 

 

 

y = x2 – 2                                           y = (x – 2)2 – 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

y = (x – 3)2 – 2                                   y = (x + 10)2 - 2

 

 

 

 

 

As you can see from these graphs, the parabola does not change shape (it does not get larger or smaller) the only thing that happens is that it changes positions.  The next graph shows this fact much better.

 

 

 

 

 

 

 

I have also included the line y = -2 to illustrate the parabolas movement.  Each of the vertices land on the line y = -2, so that means the y coordinate will always be –2.  This is easy to show by calculus or the form that the equation is in.  Let me start with the harder of the two calculus. 

 

To find the extrema of any second-degree equation you need to take the derivative. 

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Calculus Version

 

y = (x – d)2 – 2

 

y’ = 2(x – d) = 2x – 2d

 

2x – 2d = 0

 

x = d

 

Plug back into original to find y value for exterma

 

y = (d – d)2 – 2

 

y = -2

 

So the vertex of the parabola is always going to be of the form (d, -2)

 

 

 

Easy Way

 

If you look at the equation it is in the form y = (x – h)2 + k, where (h, k) is the vertex of the parabola.

Since we have y = (x – d)2 – 2, the vertex must be (d, -2).

 

 

 

 

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