I could never tell if you were Australian or English, but the rugby analogy solved the problem for me.
@user-pv5hd1vu1t
3 жыл бұрын
another way to think about f(ax - b) first translate b units to the right then with the current translated graph you have, (x , y) values turn into (x/a , y) In the context of this example: f(x) has points at (-1,-5) , (0,0) , (1,3) , (2,4) , (3,3) , (4,0) , and (5,-5) we want f(2x-6) first, translating six units to the right gives (-1 + 6,-5) , (0 + 6,0) , (1+6,3) , (2+6,4) , (3+6,3) , (4+6,0) , and (5+6,-5) (5,-5) , (6,0), (7,3) , (8,4), (9,3), (10,0), (11,-5) Then applying the dilation: (5/2,-5) , (6/2,0), (7/2,3) , (8/2,4), (9/2,3), (10/2,0), (11/2,-5) We get our points (2.5,-5), (3,0), (3.5,3), (4,4), (4.5,3), (5,0), (5.5, -5)
@TheBallzin
3 жыл бұрын
yeah thanks for joining this discussion and helping explain everything! Math is fun!
@youtubeweb3009
3 жыл бұрын
How come we half it instead of multiplying it by 2?
@user-pv5hd1vu1t
3 жыл бұрын
@@youtubeweb3009 it might be easier if you think about in terms of y-dilation. I think if you have y = 2f(x) then I think you can agree that the "y values" double. If I write it like this instead: y/2 = f(x) such that the vertical dilation factor "acts on the y-term" then you see that "reverse dilation", why am I not halving the values??? same with y = f(2x-b) then let's take the inverse 2x - b = y^-1 {note this is not 1/y, but rather the inverse function of y} 2x = y^-1 + b x = 0.5(y^-1 + b) ^if written like this, then you would see the x-values are halving indeed, but written in original form f(2x-b), you see that it has that "reverse pattern halving instead of doubling" hopefully that made it clear. a short cheat way (and possibly another explanation of why halving instead of doubling) "is just to sub in values for x and see what you get as your new output" to see if things are "halving" or "doubling" etc.... There is also a vector way of explaining things but that gets too complicated.
@wispybell
Жыл бұрын
Translating to the right 6 and compressing the function horizontally by half will result in the same graph. The common mistake there is compressing with respect to the vertex rather than the y-axis.
@yaxjoshi8655
3 жыл бұрын
lol i enjoy watching Eddie teach rather then the videos my teacher makes
@matemaatika-math
3 жыл бұрын
Where can we see the videos your teacher makes?
@hannahhackett3924
3 жыл бұрын
yh same
@yanengganak9771
3 жыл бұрын
Bro i understand now. Eddie ur the best, thanks 🔥🔥🔥🔥🔥🔥🔥🔥🔥
@TheBallzin
3 жыл бұрын
My professor uses software like statcrunch to teach our class, and then most people like barely follow along but, I see the use of software if you were doing statistics for an occupation like 24/7, but I rather learn by a calculator first, so thanks.
@henryzheng3966
2 жыл бұрын
I am confused somehow... Are we really applying the order operation to the transformations? Because it seems to me that with f(2(x-3)), the order of operations says to take care of whatever is inside the parenthesis first, but we're actually doing the horizontal shrink which is the multiplication that is outside the parenthesis. I feel like we are more like following the order of "unwrapping an expression/equation". Shifting 6 units right then shrinking half is the same as shrinking half then shifting 3 units right. Now I just don't see the point of factoring what's inside the f( ) if we just want to graph a function.
@AdamAlton
Жыл бұрын
I'm glad I'm not the only person who had this thought. It seems to me that he's confusing things here. He says that the order of operations for transforming graphs is the same as for normal calculations, but that's only true because he's factorised it (thereby changing the order of the calculations). I think it would be more accurate (and less confusing) to say: when dealing with transformations outside of the brackets (for transformations in the y-direction), the order to apply them in follows BODMAS; when dealing with transformations *inside* the function brackets (which affect the x-direction), the transformations should be applied in the opposite order (SAMDOB!); and between y-based and x-based transformations, the order doesn't matter, beccause they don't affect each other.
@anoukier
6 ай бұрын
@@AdamAlton thank you for this comment i'm pretty sure you just saved my as levels
@AdamAlton
6 ай бұрын
@@anoukier Glad to be of service :-) Good luck!
@TheBallzin
3 жыл бұрын
REMEMBER graph initially starts at y= -(x-2)^2 +4 while final position is translated to f2(x-3)
@matemaatika-math
3 жыл бұрын
What do you mean by mentioning that graph initially starts at y= -x^2 -2? That graph isn't on the picture as it'd have 2^(1/2) and -2^(1/2) as intersection positions on x-axis not 0 and 4.
@TheBallzin
3 жыл бұрын
@@matemaatika-math I guess im dumb, but i edited my first answer, to the correct initial position.
@peacecop
3 жыл бұрын
@@TheBallzin I still don't get what you want to tell with your stated initial function as it's actually y = -x^2 + 4 * x? The final position isn't translated to what you've written but it's translated using that function. The final function is -4 * x^2 + 32 * x - 60.
@TheBallzin
3 жыл бұрын
@@peacecop actually not sure how our answers are both right can you explain how you came up with your answer because I believe my answer is equivalent to yours.
@user-pv5hd1vu1t
3 жыл бұрын
@@peacecop their initial function y= -(x-2)^2 +4 and yours y = -x^2 + 4 * x are indeed equivalent. They found theirs via using vertex form. a(x-h)^2 + k
@syedziaulhaque6219
2 жыл бұрын
Your explanation is pretty good as usual however, I have a point to clarify. As for the order of transformation, shouldn’t it be opposite of bidmas as they are connected with only horizontal transformations?
@ugnemikalainyte1702
Жыл бұрын
thats what i thought too
@sebastiendamay3894
3 жыл бұрын
Hi, which tool/software do you use?
@RitobanRoyChowdhury
3 жыл бұрын
Notability, he made a video a couple months ago explaining his setup, and there's more info on his second channel Wootube2.
@alecstanleyy
3 жыл бұрын
my math teacher recommended u HAHA
@blakehowell5917
3 жыл бұрын
I need help with the integral of 2x from 10 to 13
@isakwatz11
3 жыл бұрын
First integrate 2x by answering the following question: Which function has a a derivative that is 2x? Answer is of course x^2 and you could add a constant but in this context you don’t need to. Then just calculate 13^2 - 10^2 and you are done!
@blakehowell5917
3 жыл бұрын
@@isakwatz11 I’m sorry you had a genuine reply but I was just trying to get someone to type the funny number 69 lmao
@isakwatz11
3 жыл бұрын
@@blakehowell5917 Hahahahahhaa sorry to disapoint! :)
@marshallchiasson7359
3 жыл бұрын
I think I'm really rusty because if you plug in an x value of 4 in the final equation shouldn't you get a y value of 5?
@user-pv5hd1vu1t
3 жыл бұрын
f(2x-6) at x = 4 f(2(4) - 6) = f(8 - 6) = f(2) on the original graph, f(x), f(2) = 4 thus at x = 4, y = f(2x-6) = 4 (which is not 5, so it is indeed 4 and not 5)
@matemaatika-math
3 жыл бұрын
Please show your work how you get 5.
@matemaatika-math
3 жыл бұрын
@@user-pv5hd1vu1t You seem to explain a very complated way. Why do you calculate f(2)?
@user-pv5hd1vu1t
3 жыл бұрын
@@matemaatika-math because at X=4, the "new" graph is at f(2). then proceeding back to the original function/graph, we see visually that at X=2, the corresponding y value is 4. or in other words f(X) = 4 at X=2, or f(2) =4
@peacecop
3 жыл бұрын
@@user-pv5hd1vu1t Now, I get what you mean, however you still explain a very complicated way. What I still don't get is why is f(2) important as the final abstsiss is x = 5, not x = 2. Now, I also get what the starter of this thread thought. They interchanged x and y. -4 * 5 ^ 2 + 32 * 5 - 60 === 0. And -4 * 4 ^ 2 + 32 * 4 - 60 === 4. If we want to calculate f(2) for the final function then we get -4 * 2 ^ 2 + 32 * 2 - 60 === -16.
@andrewjolly319
3 жыл бұрын
Hi Eddie, what software are you using here? I use desmos when I teach this sort of thing but I'd love to be able to write on the screen like that without having to switch back to openboard.
@techyvaibhav3516
3 жыл бұрын
he's using an ipad pro in this video
@techyvaibhav3516
3 жыл бұрын
and yea the app he's using is called notability probably
@andrewjolly319
3 жыл бұрын
Thanks!
@techyvaibhav3516
3 жыл бұрын
@@andrewjolly319 happy to help!
@jossenaert7969
3 жыл бұрын
How old are the students that this lesson is for?
@Nameless60138
3 жыл бұрын
@@abdessamad7654 hhhhh
@mrkitty777
3 жыл бұрын
12yr school kids probably
@ollieheath
3 жыл бұрын
If it helps, I'm from the UK and I was 15/16 when we learned this
@_FrozenPotato_
3 жыл бұрын
This is something you learn for the first time in Year 9 so ~15 year olds and then again as part of the topic Graphing Techniques in Year 12. So ~17 years old.
@mrkitty777
3 жыл бұрын
It depends. After learning x^2 this is a topic. 12yr old here first learn pythagoras, then parabola.
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