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.

Thankful for your teaching

You moved my phone cord when you moved =_-~~ this one

Imagine your hand grabbing something reach out and move it like a cloud

I think that's how you moved my cord

28,000 views, 4 comments? D:

Why is it that, in the first two examples, when asked for g(x) in terms of f(x) you are affecting the result of f(x) but then for the other 2, you are affecting the result of g(x) and writing it as affecting f(x)? It's like you switched the rules, that doesn't make any sense at all.

I wish khan academy videos had an intro and didnt start as abruptly.

You have super super super bad explanation why if we have graph x^2 and then we want to make new like (x+2)^2, so we should shift it to the LEFT despite on that, x+2 is always goe's to the RIGHT.

Explain this moment concretly without your favourite "just think about it" and similar phrases.

This helped me prepare for a test. Thanks a lot!

Does the last example reflect on the y or x axis

nice

I am starting transformations in algebra class tomorrow, so thanks for the early lesson and help, but PLEASE STOP REPEATING EVERYTHING YOU SAY 3 OR 4 TIMES

khan you are a genius I needed this

How can he teach everything that ever existed?

https://m.youtube.com/watch?v=5NMrrorAue8&t=34s

You are the best

I don’t like today how you explained because your title didn’t match what you talking about in this topic

2:31 It should be -3, f(-3)

"this one seems kind of wacky"-me with every math problem! 🙂