First draw sinx then sin2x and you will see why.(Original post by kiiten)
Another (different) question
If y = 1 + sinx is stretched in the xdirection, scale factor 1/2
why is it y = 1 + 2sinx
instead of
y = 1 + sin2x
??
Year 13 Maths Help Thread
Announcements

 Follow
 721
 20092016 10:07

 Follow
 722
 20092016 10:12
(Original post by kiiten)
Another (different) question
If y = 1 + sinx is stretched in the xdirection, scale factor 1/2
why is it y = 1 + 2sinx
instead of
y = 1 + sin2x
??
Sin 2x is a compression in the x direction, scale factor of 1/2, sin x/2 is a stretch of scale factor of 2 though. You can rationalise that by thinking of values for x and comparing to sin x. For example, sin(x) = 1/2 when x = 45, however sin (x/2) when x = 90 is the same as sin(x) when x = 45 and sin(x/2) therefore doesn't reach it's maxima at f(x) = 1 until x = 180. Not the easiest thing to get your head around at first and seems a bit counter intuitive, but if you can learn to understand it and explain it to yourself as it were then it becomes a lot more simple 
 Follow
 723
 20092016 10:26
(Original post by k.russell)
y = 1 + sin x isn't a stretch, it just shifts the graph of sin x up by one. So it makes it range from 0 <> 2 rather than 1 <> 1.
Sin 2x is a compression in the x direction, scale factor of 1/2, sin x/2 is a stretch of scale factor of 2 though. You can rationalise that by thinking of values for x and comparing to sin x. For example, sin(x) = 1/2 when x = 45, however sin (x/2) when x = 90 is the same as sin(x) when x = 45 and sin(x/2) therefore doesn't reach it's maxima at f(x) = 1 until x = 180. Not the easiest thing to get your head around at first and seems a bit counter intuitive, but if you can learn to understand it and explain it to yourself as it were then it becomes a lot more simple
I understand where your coming from: sin 2x is a compression and sin(x/2) is a stretch < did you mean 2sinx or is that the same thing?? 
 Follow
 724
 20092016 10:37
(Original post by kiiten)
Thanks but why is it 2sinx instead of sin2x  in my textbook it says that a stretch in the xdirection means the transformation is y=f(x) to y=f(x/c)
I understand where your coming from: sin 2x is a compression and sin(x/2) is a stretch < did you mean 2sinx or is that the same thing?? 
 Follow
 725
 20092016 11:06
(Original post by k.russell)
No!! 2sinx is a stretch in the y axis, or an increase in amplitude. This one is much easier to understand imo, if sin(x) = 1, 2sin(x) = 2 simple as that( literally the value of sin(x) multiplied by 2), so the positions of the minima and maxima are the same, but their magnitude increases and the range changes to 2 <> 2
Maybe you could help with this question:
y = sinx transformed by a sequence of transformations: translation [ 0 pi] , reflection in xaxis, stretch ydirection S.F. 4
i got y = 4sin (x  pi) which is correct.
but it can also be y=4sinx
I know i got it right but how do you get the other answer? 
 Follow
 726
 20092016 11:14
(Original post by kiiten)
So sorry to have bothered you (i made a silly mistake :P)  i looked over it again and realised i mixed up the transformations in the first place. But thank you for your help
Maybe you could help with this question:
y = sinx transformed by a sequence of transformations: translation [ 0 pi] , reflection in xaxis, stretch ydirection S.F. 4
i got y = 4sin (x  pi) which is correct.
but it can also be y=4sinx
I know i got it right but how do you get the other answer?
If you think about it, sin(xpi) shifts the graph along one half repeat to the right, so the minima now occurs at 90 degrees, and the maxima at 270. This inversion is the same as what would happen if it was sin(x), the minima > the maxima and vica versa. Then obviously they are both multiplied by 4 
 Follow
 727
 20092016 11:16
(Original post by kiiten)
So sorry to have bothered you (i made a silly mistake :P)  i looked over it again and realised i mixed up the transformations in the first place. But thank you for your help
Maybe you could help with this question:
y = sinx transformed by a sequence of transformations: translation [ 0 pi] , reflection in xaxis, stretch ydirection S.F. 4
i got y = 4sin (x  pi) which is correct.
but it can also be y=4sinx
I know i got it right but how do you get the other answer?
So . 
 Follow
 728
 20092016 11:28
(Original post by k.russell)
I don't really get the translation bit you've put in square brackets but that's not important.
If you think about it, sin(xpi) shifts the graph along one half repeat to the right, so the minima now occurs at 90 degrees, and the maxima at 270. This inversion is the same as what would happen if it was sin(x), the minima > the maxima and vica versa. Then obviously they are both multiplied by 4
I dont really get the sin(x) and multiplied by 4 part. But im not sure if i should just leave it because I dont want to confuse myself as i already got the answer right :3 
 Follow
 729
 20092016 11:30
(Original post by kiiten)
I need help on a C3 domain and range questions .... again.
Ive been over it twice and thought i understood but then got confused when i got this question wrong.
In simple terms is the domain the x value and the range the y value you get out of the function?
f(x) = (2x +5) / (x  3)
x E R, x is not equal to 3
for the first part you are told to find the range. I got it as
0< f(x)<= 13 but it should be f(x) E R, f(x) is not equal to 2.
Where have i gone wrong?? (this isnt related to the question but how is R different to x E R) 
 Follow
 730
 20092016 11:33
(Original post by kiiten)
Could anyone help me on this question before it gets buried  i still dont understand it.
For example gives which is a y value that you get out of the function, but it's not in 0 < f(x) <= 13.
Why don't you try sketching the function? You'll see that it outputs every single real number except y=2.
R is the set of real numbers. 
 Follow
 731
 20092016 11:43
(Original post by kiiten)
Could anyone help me on this question before it gets buried  i still dont understand it.
The capital E type thing means the set of, R means real numbers as Zacken said, so x E R basically just means x is a real number, like x E Z would mean x is an integer (I think) 
 Follow
 732
 20092016 11:45
(Original post by k.russell)
like x E Z would mean x is an integer (I think) 
 Follow
 733
 20092016 11:47
(Original post by kiiten)
Could anyone help me on this question before it gets buried  i still dont understand it. 
 Follow
 734
 20092016 13:26
(Original post by kiiten)
Could anyone help me on this question before it gets buried  i still dont understand it.
Okay, firstly; you need to know the general shape of graph and the fact that this has one horizontal asymptote, and one vertical asymptote. You can even see that this function goes over ALL x and y values EXCEPT those asymptote ones. Here is a graph of your function and for comparison:
Your function of is a similar to as you still have linear (function) over linear as the fraction therefore there will still essentially be ONE horizontal and ONE vertical asymptote as well. You are told the vertical one, , as is not valid due to division by 0, this is the straight forward part.
However for the horizontal asymptote, it is trickier to determine it because we need to observe what happens to as tends to positive/negative infinity, therefore I'd suggest dividing (using long division or otherwise) by and expressing it in the form which you'd find to be then consider what happens as . Of course, you can see that the second term goes to 0 therefore you are left with 2; hence the function TENDS to 2 therefore as .
It never reaches 2, therefore it cannot be equal to it. Hence where the , comes from.
EXTRA NOTE:
When it comes to finding the horizontal asymptote for this question (and any alike) you may notice that you can divide (just as you can multiply) the top and bottom of the fraction by something. In this case, you can divide top and bottom by .
You get left with .
Now if you take the limit as then you can see that AND both go to 0, and your overall fraction is as we had before.
Hope this helps! I didn't want to just plain out do the question for you, but I'd find this difficult to explain otherwise without raising even more confusion so take it with you because it will definitely help on similar questions as you're now aware of how to approach them with a solid method. This should clear up any confusion you may have about the question.
As an example, find the range and domain forLast edited by RDKGames; 20092016 at 15:33. 
 Follow
 735
 20092016 14:05
(Original post by RDKGames)
Your function of is a mere set of transformations from therefore there will still essentially be ONE horizontal and ONE vertical asymptote. 
 Follow
 736
 20092016 14:16
(Original post by Zacken)
Er... how? 
 Follow
 737
 20092016 19:29
(Original post by Zacken)
Why do you think it's 0 < f(x) <= 13
For example gives which is a y value that you get out of the function, but it's not in 0 < f(x) <= 13.
Why don't you try sketching the function? You'll see that it outputs every single real number except y=2.
R is the set of real numbers.
So whats the difference between having a domain of R and domain of x E R ? 
 Follow
 738
 20092016 19:46
(Original post by kiiten)
Ok, ill sketch it out  i usually avoid drawing graphs because i normally mess it up and get the question wrong
So whats the difference between having a domain of R and domain of x E R ?
And if you get more error from sketching graphs, then stick to the methods I provided if you're okay with them.Last edited by RDKGames; 20092016 at 19:52. 
 Follow
 739
 20092016 20:08
(Original post by RDKGames)
I'm not quite sure where you got the 0 and 13 from for the range??
Okay, firstly; you need to know the general shape of graph and the fact that this has one horizontal asymptote, and one vertical asymptote. You can even see that this function goes over ALL x and y values EXCEPT those asymptote ones. Here is a graph of your function and for comparison:
Your function of is a similar to as you still have linear (function) over linear as the fraction therefore there will still essentially be ONE horizontal and ONE vertical asymptote as well. You are told the vertical one, , as is not valid due to division by 0, this is the straight forward part.
However for the horizontal asymptote, it is trickier to determine it because we need to observe what happens to as tends to positive/negative infinity, therefore I'd suggest dividing (using long division or otherwise) by and expressing it in the form which you'd find to be then consider what happens as . Of course, you can see that the second term goes to 0 therefore you are left with 2; hence the function TENDS to 2 therefore as .
It never reaches 2, therefore it cannot be equal to it. Hence where the , comes from.
EXTRA NOTE:
When it comes to finding the horizontal asymptote for this question (and any alike) you may notice that you can divide (just as you can multiply) the top and bottom of the fraction by something. In this case, you can divide top and bottom by .
You get left with .
Now if you take the limit as then you can see that AND both go to 0, and your overall fraction is as we had before.
Hope this helps! I didn't want to just plain out do the question for you, but I'd find this difficult to explain otherwise without raising even more confusion so take it with you because it will definitely help on similar questions as you're now aware of how to approach them with a solid method. This should clear up any confusion you may have about the question.
As an example, find the range and domain for
Perhaps it would be easier if i always sketched the graph for these type of questions.
Thanks for explaining it 
 Follow
 740
 20092016 20:13
(Original post by kiiten)
>_< im struggling to understand what you're saying (i think i get the general idea though?). Just the part in bold  what do you mean by that.
Perhaps it would be easier if i always sketched the graph for these type of questions.
Thanks for explaining it
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: January 22, 2017
Share this discussion:
Tweet
Related discussions:
 The Current Year 13 Thread 20162017
 The Current Year 11 Thread Mark I (20162017)
 ***Grow Your Grades 2016****
 [18 WEEKS LEFT] 201617 Contract: What Are Your ALevel ...
 STEP Prep Thread 2017
 The Super Juicy CURRENT YEAR 11 (201213) THREAD ...
 How do you feel about the Maths forum?
 My Year 13 Journey to A*A*A*A*
 Hi.
 The Current Year 12 thread 20162017
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.