 This video will be about rational function applications. So if we read the first problem, we find out that we have this sales function that can be s is equal to 250t over t squared plus 150, represents daily sales in 10,000s. T days after the product was debuted and used the graph to answer the following. I'm assuming you can read the problem for yourself. Approximately how many days after the product came out did the sales reach a maximum? To me the maximum is right here at the top, so I look down at that and that looks like it would be about 12 days. 12 days and then we want to know what that maximum sale is. Well to do that we have to go back to our sales formula. So s of 12 is what we're really trying to find and that's equal to 250 times 12 and then 12 squared plus 150. And if we pull up your calculator and we have parentheses 250 times 12 and then all of that is divided by the bottom which says 12 squared plus 150. Press enter and we get 10.2. But remember this is in 10,000s. So we really would say that that maximum sales would be 102,000 multiplied by 10,000. To use the function to compute the rate of change for the intervals between 7 and 8 and 60 and 62 and then see what we notice. Well remember we have to find out what f or s in our case of 7 is and we want to know what s of 8 is and we want to know what s of 60 is and we want to know what s of 62 is. So go ahead and take a moment to put those into your calculator. You might just want to put this function in your calculator y equal and then look at your table since you have to have so many of them. And if we use our difference of the y's over the difference of the x's we have a difference here is going to be over 1. So it's going to be 0.55 over 1 or just 0.55. And if we do these two we get on the top when we take 4 minus 3.881 we get 0.119 and over it would be a negative 2 because we're going to take 60 minus 62. And when we do that we get a negative 0.0595 so we'll call it 0.06. That's the rate of change that we are changing per day. What does that tell us? At the beginning we were growing just like we thought but when we evened out here we really weren't growing very much. In fact it's being a negative kind of losing a little bit. Alright so use mathematical notation to state what happens to the daily sales function as the number of days becomes infinitely large. If we look at that in our table we would say that T as T goes to infinity bring back that calculator let's say like 10,000 or even bigger than that. What does our table show us? It's showing us that we are at 2.3 times 10 to the negative 4. That's 3 zeros in front of that too so we would call that close enough to be 0. So S the sales is going to go to 0. And then it says what role does the T axis play? Well if S is going to 0 as T gets very large then the T axis must be the horizontal asymptote. Alright second example. We have this cost function again. C of X is going to be the original cost of manufacturing the item. And then we've got fixed cost and then C is the actual cost and we are going to have this average. So a company develops a better disposable diaper and we see that the fixed cost and production is 20,000 per month and the cost of manufacturing the diapers is 6. By the way this in here should actually be an A of X that's a typo because here's X is on this side and a T on that side so that should be X in there. And when we have 20,000 per month that's going to be our K and when we have $6 per K so that's going to be our C above. So we want to know what the average cost function is so A of X in general would be K which is 20,000 plus our 6 times our X all over X. But we now know that X is going to be 2,000 so when we do that we end up with 32,000 over 2,000 which is just going to be 16. So the average cost function if 2,000 cases are made it would be $16. That level of production will bring the average cost down to $8 when now we know the average cost so 8 is equal to that 20,000 plus 6X over X. And if we solve both sides by X to clear the fraction so 8X is equal to 20,000 plus 6X and subtracting 6X we get 2X equal to 20,000 and if we divide by 2 we find out that that's going to be 10,000. 10,000 cases would bring the average cost down to $8 per case. And then it asks what's the significance of the horizontal asymptote for the average cost function. And this is just that the cost, this is the cost that large quantities of cases would tend toward. Cost tends toward or never gets lower, never gets lower than the vertical or horizontal asymptote. And let's figure out what the horizontal asymptote is. Well we've got up here if we look at our degrees this is degree one and this is degree one so N is equal to M. So that means it's going to be Y equal A over B so in our case it's going to be Y is equal to 6 because the coefficient would be 6 over 1. Or it's telling us that it will be never lower than $6. And finally will the company ever reach its goal of $7 per case as it's maximum in production? Well the maximum production, if we go back to the slide before, they told us that their maximum production was 16,000 cases. So that's going to be our X. So X is equal to 16,000. In fact let's change colors one more time. X is equal to 16,000 so going into the cost function of X which is actually 16,000. So we have 20,000 plus 6 times that 16,000 all over 16,000. And if you compute that you're going to end up with 7.25. So did they reach their goal? Not quite. It's 7.25 not 7.0. So not quite.