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.
