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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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