Identifying Quadratic Models

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Identifying Quadratic Models: Using y=x^2 to give a sense of why the change in the change of y (or the change in the slope) is constant

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In the last video, I told you that for a quadratic function,
that the change in the change of y is constant.
What I mean by that-- this is what we did in the last video.
This was a quadratic, this was data
from a quadratic function.
That when x increased from 1 to 2, y increased by 2.
Then when we increased x from 2 to 3, y
didn't increase at all.
It increased by 0.
So the change in y was clearly not constant.
And then we went from 3 to 4, y actually decreased by
negative 2.
So the change in y was not constant, but the change in
the changes of y-- we had a change in y of 2, then a
change in y of 0-- so our change in the change of y was
negative 2.
Then we went from a 0 to a negative 2.
So the change in the change of y was negative 2.
What I want to do in this video is show you why that
works for a quadratic function.
And just to make the math simple, I'm going to pick the
simplest of quadratic functions.
And if you have some free time on your hands, you might try
it out for the more general case.
But I'm going to show you that it works for y
is equal to x squared.
And you could try it later with ax squared
plus bx plus c.
And I'm going to increment x by 1.
This'll work as long as you increase x
by a constant amount.
But 1 makes the math a little bit simpler.
So let's pick some x's.
I'm going to keep things a little bit general.
So it's going to be very abstract and algebraicy for a
little bit.
But I think you might find it satisfying.
Then we'll figure out our y's given the x.
So let's say that x is x1.
So when I write this little subscript 1 here, that means
I'm picking a particular x, a particular example of the
variable x, a particular x in the domain of this function.
So in that situation, what's y going to be equal to?
Well, y is just going to be equal to x1 squared.
Let me do that in orange, keep the colors consistent.
x1 squared.
Now let's say we increment x by 1.
So our next one is going to be x1 plus 1.
I'm just going 1 above that number there.
If this was 3, this is going to be 4.
If this was 0, this is going to be 1.
Then this is going to be x1 plus 1 squared, which is equal
to x1 squared plus 2x1 plus 1.
All right, let's increase this by 1.
So then we get to x1 plus 2.
If this is 0, this is 1, this is 2.
If this is 10, this is 11, this is 12.
So then y, when we have this x value, will be x1 plus 2
squared, which is equal to x1 squared plus 4x1 plus 4.
Let's do two more actually.
Now let's increase by another 1.
I think you see what I'm doing.
x1 plus 3.
The y value when x is x1 plus 3 is going to
be x1 plus 3 squared.
Which is equal to x1 squared plus 6x1, plus 9.
Let's do one more.
So when x is going to be-- let's increment it by 1 more--
x1 plus 4, y is going to be x1 plus 4 squared, which is equal
to x1 squared plus 8x1, plus 16.
Now, given this, let's figure out the changes in y.
So let me write here the change in y.
The change in y as we increment x by 1 each time.
So here the change in y, we'll take this point or this y
value minus this y value.
If we take this y value minus this y value, so from here to
here, our change in y is going to be x1 squared plus 2x1,
plus 1, minus x1 squared.
So these cancel out and we just get 2x1 plus 1.
Then when we go from this point to this point, our
change in why is going to be x1 squared plus 4x1, plus 4,
minus x1 squared, minus 2x1, minus 1.
So what is this going to be equal to?
This is going to be equal to-- these guys cancel out-- 4x1
minus 2x1 is 2x1.
And then 4 minus 1 is 3.
So plus 3.
Now let's do the next one.
When I go from this y value to that y value, change in y is
going to be x1 squared plus 6x1, plus 9, minus
this guy over here.
x1 squared plus 4x1, plus 4.
These cancel out.
Sorry, I have to distribute the negative sign.
I'm subtracting this whole thing.
Put a minus there and a minus negative sign there.
6x1 minus 4x1, that is 2x1.
And then I have 9 minus 4.
So plus 5.
So this is the change in y.
The first change in y was that, the second change in y
is that, third change in y is that.
Let's do this one over here.
So our change in y is x1 squared plus 8x1, plus 16,
minus this value over here.
Minus x1 squared minus 6x1, minus 9.
I'm distributing the minus sign.
That cancels out.
8x1 minus 6x1 is 2x1.
16 minus 9 is equal to 7.
So plus 7.
So our change in y's from here to here is 2x plus 1, right
there-- 2x1 plus 1.
Then it's 2x1 plus 3.
Then it's 2x1 plus 5.
And then it's 2x1 plus 7.
So what is the change in the change of y?
Notice, these are not the same values.
They're increasing.
And so we're kind of jumping the gun.
What is the change in delta y?
So if we go from 2x plus 1 to 2x plus 3, well, if you
subtract that from that, the 2x's are going to cancel out.
3 minus 1 is 2.
If you go from 2x plus 3 to 2x plus 5, once again, the 2x's
cancel out.
We've just increased by 2.
If you go from 2x1 plus 5 to 2x1 plus 7, once again, you're
increasing by 2.
So for at least the situation of y is equal to x squared,
hopefully it satisfies you that no matter which x1 you
pick, as long as you keep increasing by 1-- and it's
actually true if you increase by any constant number-- the
change in your y values will not be constant.
Notice this is not that, which is not that,
which is not that.
But the change in the change in y's-- the change in y's are
increasing in this example at a constant rate.
So this was completely analogous to what we did here.
And if you have some time on your hands, you can actually
prove this or show it or kind of prove, you know, get
comfort that this will work for the more general case of y
is equal to ax squared or ax squared plus bx plus c.
Or it will work for ax squared plus bx plus c, if we
increment just by any arbitrary constant here.
So you could put a k, 2k, 3k.
The algebra gets a little bit hairier, that's
why I didn't do it.
But it should still work.
So hopefully you found that kind of satisfying.

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