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## Stretching functions

Current time:0:00Total duration:6:40

# Shifting & reflecting functions

CCSS Math: HSF.BF.B.3

## Video transcript

So this red curve is
the graph of f of x. And this blue curve is
the graph of g of x. And I want to try to express
g of x in terms of f of x. And so let's see
how they're related. So we pick any x. And we could start right
here at the vertex of f of x. And we see that, at least
at that point, g of x is exactly 1 higher than that. So g of 2-- I could
write this down-- g of 2 is equal to f of 2 plus 1. Let's see if that's
true for any x. So then we can just
sample over here. Let's see, f of 4
is right over here. g of 4 is one more than that. f of 6 is right here. g of 6 is 1 more than that. So it looks like if we pick
any point over here-- even though there's a little bit
of an optical illusion-- it looks like they
get closer together. They do if you look
try to find the closest distance between the two. But if you look at
vertical distance you see that it
stays a constant 1. So we can actually
generalize this. This is true for
any x. g of x is equal to f of x is
equal to f of x plus 1. Let's do a few more
examples of this. So right over here, here
is f of x in red again, and here is g of x. And so let's say we picked
x equals negative 4. This is f of negative 4. And we see g of negative
4 is 2 less than that. And we see whatever f of
x is, g of x-- no matter what x we pick-- g of x
seems to be exactly 2 less. g of x is exactly 2 less. So in this case, very
similar to the other one, g of x is going to
be equal to f of x. But instead of
adding, we're going to subtract 2 from f
of x. f of x minus 2. Let's do a few more examples. So here we have f
of x in red again. I'll label it. f of x. And here is g of x. So let's think about
it a little bit. Let's pick an
arbitrary point here. Let's say we have in red here,
this point right over there is the value of f of negative 3. This is negative 3. This is the point
negative 3, f of 3. Now g hits that same value
when x is equal to negative 1. So let's think about this. g of negative 1 is equal
to f of negative 3. And we could do that
with a bunch of points. We could see that g of 0, which
is right there-- let me do it in a color you can
see-- g of 0 is equivalent to f of negative 2. So let me write that down. g of 0 is equal to
f of negative 2. We could keep doing that. We could say g of 1,
which is right over here. This is 1. g of 1 is equal to
f of negative 1. g of 1 is equal to
f of negative 1. So I think you see
the pattern here. g of whatever is equal to the
function evaluated at 2 less than whatever is here. So we could say that g of
x is equal to f of-- well it's going to be 2 less than x. So f of x minus 2. So this is the relationship. g of x is equal
to f of x minus 2. And it's important
to realize here. When I get f of x minus 2 here--
and remember the function is being evaluated, this is the
input. x minus 2 is the input. When I subtract the 2, this
is shifting the function to the right, which is a
little bit counter-intuitive unless you go through this
exercise right over here. So g of x is equal
to f of x minus 2. If it was f of x plus 2 we
would have actually shifted f to the left. Now let's think about this one. This one seems kind of wacky. So first of all,
g of x, it almost looks like a mirror
image but it looks like it's been flattened out. So let's think of it this way. Let's take the mirror
image of what g of x is. So I'm going to try my best to
take the mirror image of it. So let's see... It gets to about
2 there, then it gets pretty close to
1 right over there. And then it gets about
right over there. So if I were to take
its mirror image, it looks something like this. Its mirror image if I were to
reflect it across the x-axis. It looks something like this. So this right over
here we would call-- so if this is g of x,
when we flip it that way, this is the negative g of x. When x equals 4, g of
x looks like it's about negative 3 and 1/2. You take the negative of
that, you get positive. I guess it should
be closer to here-- You get positive
3 and 1/2 if you were to take the
exact mirror image. So that's negative g of x. But that still doesn't get us. It looks like we
actually have to triple this value for any point. And you see it here. This gets to 2, but
we need to get to 6. This gets to 1, but
we need to get to 3. So it looks like this
red graph right over here is 3 times this graph. So this is 3 times
negative g of x, which is equal to
negative 3 g of x. So here we have f of x is equal
to negative 3 times g of x. And if we wanted to solve for
g of x, right-- g of x in terms of f of x-- we would
write, dividing both sides by negative 3, g of x is
equal to negative 1/3 f of x.