>> Let's look at an algebraic form now. We're gonna take a look at a process and we're gonna kinda build on this process. It involves function notation quite a bit. Suppose we have F of X is X squared minus 5X. Now we understand from previous work what it means to have F of X plus H. Now hang with me here. I'm going somewhere with this. But F of X plus H, it means in the F function replace X with X plus H and then evaluate, you see. So here is our function. It involves X, we replace the X with X plus H in both places. So now we have X plus H squared and we have minus 5 times X plus H. Squaring the binomial, we get X squared plus 2XH plus H squared. Now don't forget the middle term here. The tendency in going fast with problems like this is when you square a binomial, the tendency is to simply square each of the two terms and to forget about that middle term. But remember this means X plus H times X plus H, binomial times binomial. A middle term emerges in that product. Alright, and then here minus 5 times X plus H is minus 5X minus 5H. Alright, so that's the idea. Now this is just part of another more complicated form and this is called the difference quotient. And difference quotients are going to be useful as we go on with our discussions in another chapter. But it requires that we take a function like this, F function up here and we apply the difference quotient to that function. And all we're doing is kinda following our nose and following the idea of replacement within functions. You see, F of X plus H is what we did right here. And it's a blind replacement at least in the first step because we're taking the F function and everywhere we see X we are replacing with X plus H, you see, and that's what we did right here. And then minus F of X, well F of X is just the function we started with. So we're gonna subtract that function. Now it's important that if that function involves more than one term that we introduce a parenthesis because we're subtracting everything in the function, you see. So we have to show minus everybody in the function and then all of this is over a simple H. Alright, let's go through it. F of X plus H, now let's do the replacement from the top. It means in the F function replace X with X plus H. Now the original function was X squared minus 5X. So instead of X squared we're gonna have X plus H squared. And then the other term is minus 5X, we're gonna replace that X with X plus H, you see, just as we did up here. But it's a part of a bigger picture than minus F of X. So we wanna subtract the function itself. So we're subtracting X squared minus 5X. And again the introduction of the parenthesis to make sure that we subtract all of the terms involved in the function and all of these is our range. Now, then we expand in the numerator and collect to simplify X plus H squared is X squared plus 2XH plus H squared minus 5 times X plus H minus 5X minus 5H. And then minus sign, preceding the parenthesis means we changed both sides, minus X squared plus 5X. And some good things usually happen in this situation. And here we have this X squared and this minus X squared go out, and we have minus 5X plus 5X so those terms go out. And now we're left with 1, 2, 3 terms over H and I have repeated those 3 terms up here. And I noticed that they all involve H. So I can snag H out of those terms as a common factor as I've done here and then I can cancel. Now, and that's the safe way to do it. A rather unsafe way is to try to cancel this H with one of those with this one and with that one. It's possible to do that but you have to keep track of the fact that you have to divide H in every term, you see, and it's a lot safer just to extract the H in the numerator and then show the cancelation situation. And now we have the H plus 2X minus 5 which we could manipulate in a number of different ways.