Modeling ticket fines with exponential function | Algebra II | Khan Academy

Sarah Swift got a speeding ticket on her way home from work. If she pays her fine now, there will be no added penalty. If she delays her payment, then a penalty will be assessed for the number of months t that she delays paying her fine. Her total fine f in euros is indicated in the table below.

These numbers represent an exponential function, so they give us the number of months that the payment is delayed and then the amount of the fine. This is essentially data points from an exponential function.

Just to remind ourselves what an exponential function would look like: this tells us that our fine, as a function of the months delayed, is going to be equal to some number times some common ratio to the t power.

This exponential function is essentially telling us that our function is going to have this form right over here. So, let's see if we can answer their questions.

The first question is: what is the common ratio of consecutive values of f? The reason why r, right over here, is called the common ratio is that it's the ratio that, if you look at any two values, say if you were to increment t by one, the ratio of that to f of t should be consistent for any t.

Let me give you an example here: the ratio of f of 2 to f of 1 should be equal to the ratio of f of 3 to f of 2, which should be the same as the ratio of f of 4 to f of 3 or, in general terms, the ratio of f of t plus 1 to f of t should be equal to all of these things. That would be the common ratio.

So, let's see what that is. If we just look at the form, what's the ratio of f of 2 to f of 1? 450 divided by 300, well that's 1.5. That's 1.5. 675 divided by 450, that's 1.5. 1012 divided by 675, that's 1.5. So, the common ratio in all of these situations is 1.5.

So, the common ratio over here is 1.5. Just to make it clear why this r, right over here, is called the common ratio, let's just do this general form.

So, f of t plus 1, well that's just going to be a times r to the t plus 1 power, and f of t is a times r to the t power. So what is this going to be? This is going to be, let's see, this is going to be r to the t plus 1 minus t, which is just going to be equal to r to the first power, which is just equal to r.

So this term, this variable r, is going to be equal to this common ratio. When we figured out that the common ratio is 1.5, that tells us that our function is going to be of the form: f of t is equal to a times instead of writing an r there, we now know that r is 1.5 to the t power.

Now let's write a formula for this function. Well, we've almost done that, but we haven't figured out what a is. To figure out what a is, we could just substitute. We know what f of 1 is; when t is equal to 1, f is equal to 300. So we should be able to use that information to solve for a.

We could have used any of these data points to solve for a, but let's do that. f of 1 is equal to a times 1.5 to the first power, or a times 1.5, and that is going to be equal to… they tell us that f of 1 is equal to 300.

Another way of writing this is: we could say 1.5 times a is equal to 300. Divide both sides by 1.5, and we get a is equal to 200. Therefore, our formula for our function is going to be 200 times 1.5 to the t power.

Now, what is the fine in euros for Sarah's speeding ticket if she pays it on time? Paying it on time implies that t is equal to zero, or another way of thinking about it, we need to figure out her fine for t equals zero, so we need to figure out f of zero.

So what's f of zero? It’s 200 times 1.5 to the zero power. 1.5 to the zero power is one, so that’s just going to be equal to 200 euros.

Now, another way of thinking about it is: look, let's look at the common ratio. To go from 675 to 450, you're essentially dividing by the common ratio. To go from 450 to 300, you are dividing by the common ratio.

So then, to go from t equals one to t equals zero, you would divide by the common ratio again and you would get 200. Or another way of thinking about it is every time we are multiplying by the common ratio.