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## Heart of algebra

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# Interpreting linear functions — Basic example

## Video transcript

- The amount of money that
farmers in Massachusetts paid to maintain their
crops between 1991 and 2008 is modeled by the equation above, where P is the amount of
money the farmers paid, in millions of dollars, and t is the year. So, this is how much they
paid in millions of dollars, t is the year, but they're saying assuming 1991 is t equals zero. What does the 3.53 mean in the equation? So, let's look at this. So, in 1991 when t is equal to zero, this whole term is going
to be equal to zero, and the farmers are going to pay, P is going to be a hundred, so they're gonna pay
hundred million dollars in 1991 to maintain their crop. These are all the farmers in Massachusets. Now, as t increments, each
time t increases by one, the amount that the farmers
pay is going to increase by 3.53 times 1. So, one way to think about
it is this is the rate of increase from year to year. As t goes up a year, the
amount the farmers pay is going to increase by
3.53 million dollars. So, let's see which of these choices are consistent with what I just said. (laughs) The cost for maintaining crops was $3.53 million dollars in 1991. No, that's just not true. In 1991, this term is zero, and it was a hundred million dollars. The cost for maintaining crops
was $3.53 million in 2008. No, that's not going to be true either, because it's a hundred million in 1991, and then each year, it's going to increase by 3.53 million. The cost for maintaining crops increased a totally of $3.53 million
between 1991 and 2008. No, it's going to increase
$3.53 million per year, not over the entire time span. The cost for maintaining
crops increased by $3.53 million each year between 1991 and 2008. That is exactly right. Every time t, and we go forward a year, t increases by one. It's going to increase, we're going to have 3.53
times that one higher t, so, we're gonna increase
the whole P by 3.53. That kind of rhymed.