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)
@@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.
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.
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.
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.
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.
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?
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.
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!
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 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
@@沈博智-x5y 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.
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 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.
@@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.
@@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
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.
