About this video
Sal compares pairs of graphs and writes g(x) in terms of f(x). He shows vertical shifts (adding or subtracting outside f), a horizontal shift (g(x) = f(x - 2) moves the graph right), and a reflection with vertical stretch (f(x) = -3g(x), so g(x) = -1/3 f(x)).
You'll be able to
- recognize a vertical shift and write it as g(x) = f(x) + k.
- recognize a horizontal shift and explain why f(x - 2) moves a graph to the right.
- identify a reflection across the x-axis combined with a vertical stretch or squash.
- test a transformation by comparing matching points on two graphs.
Key words
- vertical shift
- Moving a whole graph up or down by adding or subtracting a number outside the function.
- horizontal shift
- Moving a whole graph left or right by changing the input, as in f(x - 2).
- reflection across the x-axis
- Flipping a graph upside down by multiplying the outputs by -1.
- vertex
- The turning point of a curve such as a parabola.
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.
Video © Khan Academy, used under CC BY-NC-SA. Watch on Khan Academy. Puzzicle is not affiliated with Khan Academy.