Hedwigeeeee
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sorry, I cannot attach a photo...there is a question about math:
a curve with equation y=sin(ax-b),where a>0,0<b<pi,the curve cuts x-axis at P,Q,R as shown, givenP(pi/10,0),R(3pi/5,0),R(11pi/10,0),find a and b.
the answer is a=2,b=pi/5, and they made a simultaneous equation :a pi/10-b=0 and a3pi/5-b=0,and got the answers.
but I wonder if a and b represents change of graph, then they should follow the role 'f(ax+b) means the x-coordinate of the points divided by a and minus b,which is opposite...I am confused why the answer just simply substitute the coordinates into ax-b,without times a becomes divided by a,minus b become plus b.
I am really thankful if anyone could explain to me about this
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RDKGames
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(Original post by Wuxinyu)
sorry, I cannot attach a photo...there is a question about math:
a curve with equation y=sin(ax-b),where a>0,0<b<pi,the curve cuts x-axis at P,Q,R as shown, givenP(pi/10,0),R(3pi/5,0),R(11pi/10,0),find a and b.
the answer is a=2,b=pi/5, and they made a simultaneous equation :a pi/10-b=0 and a3pi/5-b=0,and got the answers.
but I wonder if a and b represents change of graph, then they should follow the role 'f(ax+b) means the x-coordinate of the points divided by a and minus b,which is opposite...I am confused why the answer just simply substitute the coordinates into ax-b,without times a becomes divided by a,minus b become plus b.
I am really thankful if anyone could explain to me about this
What is the diagram they post? Also, for a3pi/5-b=0 do you really mean for it to be zero on the RHS? [Never mind, I have found an image online]


Note that \sin(ax-b) is a transformation of \sin x. It represents a horizontal translation to the right by b units, followed by a horizontal stretch with scale factor \dfrac{1}{a}. (See spoiler below for why)

Notice that P,Q,R are the first three consecutive roots of \sin(ax-b), whereas the first three consecutive roots of \sin x are 0,\pi,2\pi.

So, you can approach this problem by imagining that transforms into \dfrac{\pi}{10}, and then \pi transforms into \dfrac{3\pi}{5}, and 2\pi transforms into \dfrac{11\pi}{10}.

Take the first case. We begin with 0 and the first step is a stretch to the right by b units. This means 0 \mapsto (0+b) and this is our new coordinate. Then we stretch by scale factor \dfrac{1}{a} and obtain (0+b) \mapsto \dfrac{0+b}{a}.

Thus, this is our final new coordinate, and it corresponds to \dfrac{\pi}{10}.

Hence, \dfrac{\pi}{10} = \dfrac{0+b}{a} and you can rearrange this into \boxed{a\dfrac{\pi}{10} - b = 0}.

Repeating similarly by transforming \pi to \dfrac{3\pi}{5} yields \boxed{a\dfrac{3\pi}{5}-b = \pi}.



Spoiler:
Show

Consider the function y=f(x) being transformed into y=f(ax-b). We aim to understand why this is precisely a horizontal shift to the right by b units followed by a horizontal stretch with scale factor 1/a.

Take a single point on y=f(x), let's say (X,Y) is a point. Hence, the equation Y=f(X) is satisfied.

Now, we seek the transformed point X' for which (X',Y) lies on our *new* curve. Hence, the equation Y = f(aX'-b) (*) is satisfied. Note that we can impose the condition that aX'-b = X (**), because then f(aX'-b) = f(X) = Y hence the equation (*) certainly holds! So, rearrange (**) for X' to get X' = \dfrac{X+b}{a}.

So what we have found is that our new coordinate X' in relation to the old coordinate X is simply a shift to the right by b units (represented by +b) and then stretched by factor \dfrac{1}{a}.

Hence, f(ax-b) takes on values values x which have been shifted to the right by b units, and stretched by factor 1/a as compared to the original function f(x).
Last edited by RDKGames; 10 months ago
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Hedwigeeeee
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thank you, but I cannot understand why f(ax+b)means b units to the left,why not right, I just treated it as a rule before and do not understand it actually.
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Me2Peeps
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(Original post by Wuxinyu)
thank you, but I cannot understand why f(ax+b)means b units to the left,why not right, I just treated it as a rule before and do not understand it actually.
It is given that b lies b/w 0 and pi. So it is a positive integer. Therefore, the graph shifts to the left.

To make it easier to understand, draw the graph of y =x, y=x+1 and y=x-!. You will see what I'm talking about...
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Hedwigeeeee
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(Original post by RDKGames)
What is the diagram they post? Also, for a3pi/5-b=0 do you really mean for it to be zero on the RHS?
yes, I do, but why it must be (0+b)/a,not 0/a+b to strech first then shift?
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RDKGames
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(Original post by Wuxinyu)
yes, I do
No you don't I have just checked the mark scheme and they do indeed agree with me. It should be \pi on the RHS, not zero.

but why it must be (0+b)/a,not 0/a+b to strech first then shift?
Order of transformations matters. I believe there is a chapter in your course about this which you should cover. This is not a teaching place so I'm not going to explain it to you, especially after I have just explained the transformations above in the spoiler. Just take it at my word for now until you cover it. Or look online for videos which can teach you about it.

f(x) being shifted, then stretched yields f(ax-b).

f(x) being stretched, then shifted yield f(a(x-b)).

So there is a difference!
Last edited by RDKGames; 10 months ago
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RDKGames
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(Original post by Wuxinyu)
thank you, but I cannot understand why f(ax+b)means b units to the left,why not right
See spoiler in my first post.
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Hedwigeeeee
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ok,thanks a lot
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