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CCSS Math: HSF.BF.A.2, HSF.LE.A.2

So we see here in this table, that for given "n's", when "n" is one, "f" of "n" is 12, when
"n" is 2, "f" of "n" is 5, and when "n" is three, "f"
of "n" is negative two, when "n" is four, "f"
of "n" is negative nine. And so one way to think about it is this function "f" is defining a sequence where the first term
of this sequence is 12. The second term of this sequence is five. The third term of this
sequence is negative two. The fourth term of the
sequence is negative nine. And it goes on, and on, and on. And you might notice that
it's a arithmetic sequence. We start with a 12, and
then the next term... What have we done? We've subtracted seven. Now to go from the second
to the third term... What do we do? We subtract seven. So each term is seven less
than the term before it. Now with that out of the
way, see if you can define this function of "n." If you
can define it explicitly. So figure out a function definition. So I want to figure out
"f" of "n" is equal. I want you to figure out
what this needs to be so that if you input "n" here, it gives you the appropriate "f" of "n." So let's think about it a little bit. It's going to be, we could think
of it as we're starting at, The first term is going to be 12. But then, we are going to subtract seven. And what are we going to subtract seven? How many times are we
going to subtract seven? So for the first term, we
subtract seven zero times. And so we just get 12. For the second term,
we subtract seven once. For the third term, we subtract
seven twice. One, two times. For the fourth term, we
subtract seven three times. So it looks like whatever term we're on, we're subtracting seven "n" minus one, we're subtracting seven
whatever term we're on, that term minus one time. So it's "n" minus one times. And let's see if this actually works out. So "f" of one is going to
be 12 minus seven times one minus one, that's a zero. So that's all just going to be 12. "f" of two is going to be 12 minus seven times two minus one. So it's going to be 12
minus seven times one. We're just going to subtract seven once, which is exactly the case. We started at 12, we subtract seven once. "f" of three, you can keep testing this. 12 minus, and we should have
to subtract seven twice. And we see three minus one, is two times. So we're going to
subtract seven two times. So this looks right on. We've defined the function explicitly. We've defined "f" explicitly
for this sequence. Let's do another example, here. So in this case, we have some function definitions already here. So you have your sequence, it's kind of viewed in this table. You could view it as the
first term is negative 100. The next term is negative
50, next term is zero, next term is 50. And it's very clear that this is also an arithmetic sequence. We're starting at negative
100, and then, we're adding 50. And then we're adding 50, and then we are adding 50. So each term is 50 more
than the term before it. And what I want you to
do is pause the video, and think about which
of these definitions of the function "f" are correct. And it might be more than one. Alright, so let's think about it. So this definition right over here, one way to think about it is by saying I'm going to start at negative 100. And I'm going to add
50, "n" minus one times. Does this make sense? Well, for the first term,
if we start at negative 100, we don't want to add 50 at all, we want to add 50 zero
times, and it works out. Because one minus one is going to be zero. So it checks out for "n" equals one. Let's see, for "n" equals two,
you start at negative 100, I want to add 50 once. So this should be a one. Two minus one, yup, it's a one! We're adding 50. Whatever this number is, whatever "n" is, we're adding 50 one less that number of times. So for here, we're adding 50 twice. When "n" is four, we're
adding 50 three times. And this one checks out. When "n" is four, we're adding 50, four minus one, three times. Negative 100, plus 50 times three. We're adding 50 three times, adding 50 one, two, three times. Well that gives you
50. So I like this one. Now let's see about this one over here. Negative 150 plus 50 "n." Alright, that's one way
of saying, so let's see If "n" is equal to one,
it's going to be negative... Actually, let me draw
a table for this one. So if we have "n" and we have "f" of "n." This is going to be for this
character right over here. So if "n" is one, it's going
to negative 150, plus 50. Which is negative 100,
yup that checks out. When "n" is two, we get negative 150, plus 50 times two, which is going to be... This is 100, and there's negative 150, this is going to be negative 50. When "n" is three, and
that checks out, of course. When "n" is three, you get negative 150, plus 50, times three,
which is equal to zero. This checks out. This one over here is going to work. And you might say, "Well, hey, these formulas look different." Well you can algebraically manipulate them to see that they're the exact same thing. If you were to take this first one, it's negative 100, plus,
let's distribute this 50, plus 50 "n", plus 50 "n", minus 50. Well, negative 100, minus
50, that's negative 150. And then you have plus 50 "n." So these are algebraically the exact same definition for our function. Now what about this one here? Negative 100 plus 50
"n", does this one work? Let's see, when "n" is equal to one, this would be negative 100
plus 50, which is negative 50. Well no, this doesn't work. We need to get a negative 100 here. So this one is not, not correct.