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## Integrated math 3

### Course: Integrated math 3>Unit 6

Lesson 4: Scaling functions

# Scaling functions horizontally: examples

The function f(k⋅x) is a horizontal scaling of f. See multiple examples of how we relate the two functions and their graphs, and determine the value of k.

## Want to join the conversation?

• Transformations of functions is the most trickier and interesting topic I've seen since joining khan academy. Scaling vertically and horizontally have connection, don't they ? if we scale by the same factor, are they the same in the linear function y=x and different in y=x^2
• Its just so round about, it does not make too much sense.
I am going to attempt to explain this just so I can better understand it.

g(x) = f(2x)
means: in this case the input of g is twice as big as input of f or f input has to be multiplied by two for it to equal g input. This is the same as g(x/2) = f(x) (we have to divide input of g by two just for it to equal input of f)

In summary, if we want to graph g, we should explain what g does to make more intuitive sense.
Since g(x/2) = f(x), then g does the following.
g(takes in input of f and divide it by two)

so
g(f(-4/2)) = (x = -2, y = is the same)
g(f(4/2)) = (x = 2, y = is the same)

since we are only changing x (input) and not touching the f(x) (output) we just keep the output the same for g(x) and f(x).
• I know what you mean. I couldn't get this at all. What I just did was that when g(x) = f(2x) it makes it thinner, and when g(x) = f(x/2) then it makes it wider. That doesn't make any sense either, but it's the best I could do. I don't understand this lesson at all.
• I wanna share this method which I came up after rewatching the video over and over again.(got the idea from Sal's vid)

I was so confused when I first watched this video, then today I attempted to understand this video once again. But I still think Sal's way of explanation was a bit rough(especially problem1). I think I would have gotten more confused if I used his method to do the problems in the exercise. So came up the method similar to Sal's but uses different angle to see the problem, and I also made it more understandable for myself. Feel free to correct me if I have made any mistakes.

>First problem in the video:

Sal uses table to solve it, but the part that confused me is how he picks the x in the table. He just put 0 2 -2 (x from g(x) seemingly out of nowhere. Yes, we know the relationship between the two functions is g(x)=f(2 x), which means two times of x would give us the f's point which we can go to the graph to refer it and find the corresponding point on y-axis f(2x). So two times x, 2⋅2=4, f(4) which we can see on the graph is equal to 0. But how do you find a appropriate x at first glance, so that we can use that to refer points on graph?? What if the question gives us more complicated function like cube root or x to fifth power instead of 2x? There's no way you can figure them out that fast. Not trying to disrespect Sal, he is a amazing teacher. But in this particular lesson, his method just seems straight out copy from textbook, he knows the answer from the start, and use the answer to explain how you solve it. ( at least thats how it looks to me) Which is like putting cart before horses.

My method(i feel more comfortable with): Unlike how Sal does in the video, where he figures out x of g(x) right on the get-go. My ways of doing is first find the points we can refer to on the graph, f(x)=?, then slowly figure out the corresponding points of g(x). Essentially its just reversed of Sal does in the video.

①We can clearly see the dotted points on the graph indicate what the f(x) and x are.

[from left to right]

f(-4)=0

f(0)=g(0)≈ 5.2 !we dont need to know the exact value for it, but all we need to know is that f(0)=g(0) would be on the same point. △(I will explain the details below.)

f(4)=0

②Base on this points we can figure out the x from g(x) using the relationship g(x)=f(2x). And use some simple algebra to figure out the x of f(2x) we picked from on the graph.

first point
g(x)=f(2x)=f(-4)=0

☆Focus on f(2⋅ -2)=f(-4), ignore the f for now, and rewrite them as 2x=-4(divide both side by 2) x=-2 ,then substitute back the x

g(-2)=f(2 ⋅ -2)=f(-4)=0 simplify them
g(-2)=f(-4)=0
g(-2)=0 , f(-4)=0 there you have it, first point on g function is done. (-2,0)〆

③one down two to go, repeat the same process to figure out the rest of g(x)'s point :

second point
g(x)=f(2x)=f(0) △≈5.2
☆Focus on f(2x)=f(0), ignore the f for now, and rewrite them as 2x=0(divide both side by 2) x=0 ,then substitute back the x

g(0)=f(2 ⋅ 0)=f(0)=0 simplify them
g(0)=f(0) △because the two functions share the same points on both x-axis and y-axis, they overlap each other. You just need to plot it on the same point as f(0)〆

third point
g(x)=f(2x)=f(4)=0
☆Focus on f(2x)=f(4), ignore the f for now, and rewrite them as 2x=4(divide both side by 2) x=2 ,then substitute back the x

g(2)=f(2 ⋅ 2)=f(4)=0 simplify them
g(2)=f(4)=0
g(2)=0 , f(4)=0 there you have it third point on g function is done. (2,0)〆

④Now we have all three points〆 we can plot them on the graph, and the rest you just graph it like in the video.

`Put it all together the solutions should be like this:( figured out f(-4)=0 ,f(0)=g(0), from the graph)g(x)=f(2x)=f(-4)=02x=-4 x=-2g(-2)=f(-4)=0 therefore first point of g is (-2,0)g(x)=f(2x)=f(0)2x=0 x=0g(0)=f(0) the two function share the same points on both x-axis and y-axis, therefore they overlapped each other. (In all honesty, this process can be omitted if you know how it works)g(x)=f(2x)=f(4)=02x=4 x=2g(2)=f(4)=0 therefore third point of g is (2,0)plots all three points and connect them.`

It might seem like i overcomplicate things, but you will see this would work for most of the questions you see in the exercise. You dont have to sweating to figure out the points just like i do using Sal's method.

>Second problem from the video:

Not much different from first problem, except the relationship between function f and g is not given. But you can figure them out just like what Sal does. Second problem Sal explains much clearer than the first problem, he picks 2 , 4 , 7 from function f on the graph, which is similar to what my method or rather I took inspiration of his ways of solving and put a spin on it.

①Before anything else we have to figure out the relationship first. We know from the question that g(x)= (x/2 -4)^2 -4 and f(x)=(x-4)^2-4 , notice how two equation are almost equivalent, except for x? We can transform that to the other without needing to change whole equation. Just replace x in f(x) to x/2. f(x) becomes f(x/2) and substitute it we will get the same equation as g(x)

[x/2=1/2 ⋅ x]

f(x/2)= (x/2 -4)^2 -4 =g(x) *transitive property

f(x/2)=g(x) , g(x)=f(x/2)

Now we know the relationship between f and g. We find figure out the points of g(x) which can help us graph the function.

②Same process as I have Illustrated in problem one
( figured out f(2)=0 ,f(4)=-4, f(6)=0 from the graph)

`g(x)=f(x/2)=f(2)=0x/2=2 x=4 (times 2 both sides)g(4)=f(2)=0 therefore first point of g is (4,0)g(x)=f(x/2)=f(4)=-4x/2=4 x=8g(8)=f(4)=-4 therefore second point of g is (8,-4)g(x)=f(x/2)=f(6)=0x/2=6 x=12g(12)=f(6)=0 therefore third point of g is (12,0)plots all three points and connect them`.

>problem 3: I do it the same way as Sal, I dont have any issue with it.

Hopefully, this helps anyone who have troubles understanding the video.
• This is so confusing, I wish they had made summary/clarification/FAQ for this particular lesson like what they do for the lessons previously.
• Does anyone understand how he got the x values for the table? Why are the 1/2x values the one used in the graph? Why aren't those just the x values?
• Explanation 1:
Look carefully at both function f(x) and g(x).

You will see that the only difference between them is that in g(x), x is multiplied by (1/2).

So, g(x) = f(x/2).

Explanation 2:
f(x) = (x - 4)^2 - 4 ---- {Given in the question).
Define a new variable b. Let b = (x/2).
Substitute b into function f.
f(b) = (b - 4)^2 - 4
Since b = x/2.
f((x/2)) = ((x/2) - 4)^2 - 4
This is the exact same function as g(x).

----------------------------------------

He used those x values because they have an easy to determine y value.

Hope this helps.
• Where did he get g(x)= f(0.5x)? It looks like he just ignored the whole rest of the equation.
• He got g(x) = f(0.5x) from the first function of the graph of f(x). If you look at both of the equations of f(x) and g(x) you will notice that they both have the same horizontal translation and vertical translation than that of the parent function of x^2. The only change is that g(x) is a horizontal stretch by a factor of 2 than f(x). Thus he ignored the rest part of the equation since that was not required for graphing. If by any chance the graph of g(x) was to be graphed on the basic of the parent function then, yes, all of the characteristics of the graph needs to be in mind. But in this example you are graphing g(x) on the basis of f(x) so doing the translation which has been done already will lead to incorrect results. That is why he graphed g(x) = f(0.5x) rather than graphing the whole equation again.

Hope this helps
• why is it 3x and not x/3 for g(x)?
• we know when:

f(-3) = g(-1)
f(6) = g(2)

So the input of f is always 3 times the input for g.
So if the input for g is x, then the input for f has to 3x.
Hopefully that makes sense!