 As we discussed in the previous video, you can find the average rate of change of a function by using this formula f of x2 minus f of x1 over x2 minus x1. And remember, this is essentially the same as finding the slope of a line by finding the rise over the run. Well, the difference quotient is also another method of finding the average rate of change. But rather than getting simply one number, we're getting an equation based on how big the interval is. So we're just going to relabel our points here. In this drawing, say that this curved line is f of x and this time we know our starting point A, as labeled here, is x, so the y value would be f of x, and then our second point B, rather than knowing the coordinates of it, what we do know is we know the length of the interval. So we know we're going to have an interval of h in between our two x values. So that makes this new x value x plus h. And then our output is f of x plus h. So what that gives us if we do the rise over the run, well, the rise here, again, is our change in y values. So f of x plus h minus f of x. And the run is given to us already. It's h. So our formula for the difference quotient is simply f of x plus h minus f of x, all divided by h. Now let's take a look at an example of how you can actually apply this to an equation. So the first equation we're going to look at is a basic linear function f of x equals 5x minus 6, and we're just going to find the difference quotient for this. So the difference quotient formula, as we said before, begins with f of x plus h. So the first thing that you'll want to do is evaluate f of x plus h. What that means is in your original equation, you need to replace all of the x's with x plus h and then simplify as much as possible. So I substituted in x plus h, and now I am expanding to get 5x plus 5h minus 6. Then the second thing in our difference quotient formula was f of x, which here is already given to us. So when the difference quotient asks to do f of x plus h minus f of x, in this instance that would give us, I'll put f of x plus h in brackets, minus 5x minus 6. And then the final step for our difference quotient is to divide everything by h. So what we just need to do here is simplify the numerator, and if you distribute this negative sign to each piece, that might help you to combine like terms. But what should always happen is this original equation should cancel out with terms in your f of x plus h. So you'll notice the negative 6 and positive 6 and the negative 5x and positive 5x cancel. So in the numerator we're left with 5h, in the denominator we're left with h. And if we cancel out the h's, our difference quotient here is 5. So the question here, what does a difference quotient of 5 mean in this case? Well the difference quotient gives us a formula for the average rate of change. And what that means in this case is for our function that we started with, it means to find the average rate of change on any interval we can use the formula y equals 5. Well y equals 5 makes sense because here we have a line and so the average rate of change is going to be constant because it's a line, the average rate of change will equal what our slope was. Okay, let's look at one more example where it's not a constant average rate of change. I'm going to look at the example f of x equals x squared. Next I'm going to find the average rate of change by finding f of x plus h. So I replace the x with x plus h and remember x plus h squared really means we have to do x plus h times x plus h. So if you have to foil that out, feel free to do so. I've done it so many times it's always equal to the same thing. I know that it will equal x squared plus 2xh plus h squared. So that is my f of x plus h. So in my difference quotient formula that is the first piece. Now the second part of the difference quotient is to subtract f of x which is just the original function and then the final thing is to divide everything by h. So if you look here in our top, this negative sign gets distributed to this negative x squared and so we have a positive x squared which cancels out a negative x squared. Okay, now what we're left with, I'm going to go to a new slide here. So what we are left with is 2xh plus h squared all over h. And what you should be able to do now is cancel out an h from everything in the top. If you need to you can think about it in terms of factoring. See if you can factor an h out of everything in the numerator and then that will cancel with the h in the denominator. So what we are left with here is our difference quotient, 2x plus h. Now how do you use that to find the average rate of change? Let's say that we want to find the average rate of change on the interval from 1 to 4. Well this first value here is always our x, what we're starting with. So in this formula we'll plug in a 1 for our x, but the h, if you remember back to a few slides ago, the h is the length of our interval. So the way to find the h will be to take 4 minus 1 which is 3. And so we can substitute in the x of 1 and h of 3 into our difference quotient formula and we know that now the average rate of change on this interval from 1 to 4 is 5.