Year 13 Maths Help Thread

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    (Original post by kiiten)
    Another (different) question

    If y = 1 + sinx is stretched in the x-direction, scale factor 1/2
    why is it y = 1 + 2sinx
    instead of
    y = 1 + sin2x

    ??
    First draw sinx then sin2x and you will see why.
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    (Original post by kiiten)
    Another (different) question

    If y = 1 + sinx is stretched in the x-direction, scale factor 1/2
    why is it y = 1 + 2sinx
    instead of
    y = 1 + sin2x

    ??
    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
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    (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
    Thanks but why is it 2sinx instead of sin2x - in my textbook it says that a stretch in the x-direction 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??
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    (Original post by kiiten)
    Thanks but why is it 2sinx instead of sin2x - in my textbook it says that a stretch in the x-direction 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??
    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
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    (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
    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 x-axis, stretch y-direction 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?
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    (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 x-axis, stretch y-direction 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?
    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(x-pi) 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
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    (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 x-axis, stretch y-direction 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?
    You can also just use trig identities, you know that \sin(x-\pi) = \sin x \cos \pi - \sin \pi \cos x = -\sin x

    So -4\sin(x-\pi) = -4(-\sin x) = 4\sin x.
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    (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(x-pi) 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
    (Original post by Zacken)
    You can also just use trig identities, you know that \sin(x-\pi) = \sin x \cos \pi - \sin \pi \cos x = -\sin x

    So -4\sin(x-\pi) = -4(-\sin x) = 4\sin x.
    Thanks - that makes sense but i havent come across that trig identity so ill leave it for now.

    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
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    (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)
    Could anyone help me on this question before it gets buried - i still dont understand it.
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    (Original post by kiiten)
    Could anyone help me on this question before it gets buried - i still dont understand it.
    Why do you think it's 0 < f(x) <= 13

    For example x=1 gives f(1) = \frac{7}{-2} = -3.5 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.
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    (Original post by kiiten)
    Could anyone help me on this question before it gets buried - i still dont understand it.
    Zackens answer is good and you should heed his advice w.r.t. the function, 1/x type functions have an asymptote on both axes so finding the asymptote is basically finding the range.
    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)
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    (Original post by k.russell)
    like x E Z would mean x is an integer (I think)
    Yes, that's correct.
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    (Original post by kiiten)
    Could anyone help me on this question before it gets buried - i still dont understand it.
    You may want to read this: http://www.thestudentroom.co.uk/show...5#post63523885 and this http://www.thestudentroom.co.uk/show...9#post56184069
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    (Original post by kiiten)
    Could anyone help me on this question before it gets buried - i still dont understand it.
    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 g(x)=\frac{1}{x} 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 g(x) for comparison:

    Name:  asdfasdfsadf.PNG
Views: 16
Size:  15.9 KB


    Your function of f(x)=\frac{2x+5}{x-3} is a similar to g(x) 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, x=3, as f(3) 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 f(x) as x tends to positive/negative infinity, therefore I'd suggest dividing (using long division or otherwise) 2x+5 by x-3 and expressing it in the form A+\frac{B}{x-3} which you'd find to be f(x)=2+\frac{11}{x-3} then consider what happens as x\rightarrow \pm \infty. 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 f(x) \rightarrow 2 as x\rightarrow \pm \infty .

    It never reaches 2, therefore it cannot be equal to it. Hence where the f(x) \in \mathbb{R}, f(x) \not= 2 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 x.

    You get left with \displaystyle f(x)=\frac{2+ \frac{5}{x} }{ 1- \frac{3}{x} }.

    Now if you take the limit as x \rightarrow \pm \infty then you can see that \frac{5}{x} AND \frac{3}{x} both go to 0, and your overall fraction is f(x) \rightarrow \frac{2}{1} \Rightarrow f(x) \rightarrow 2 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 \displaystyle h(x)=\frac{3x-5}{2x+1}
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    (Original post by RDKGames)
    Your function of f(x)=\frac{2x+5}{x-3} is a mere set of transformations from g(x) therefore there will still essentially be ONE horizontal and ONE vertical asymptote.
    Er... how?
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    (Original post by Zacken)
    Er... how?
    Whoops, made a mistake in my working. Will edit that part.
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    (Original post by Zacken)
    Why do you think it's 0 < f(x) <= 13

    For example x=1 gives f(1) = \frac{7}{-2} = -3.5 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.
    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 ?
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    (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 ?
    Means the same thing essentially. Having domain of \mathbb{R} just means the domain is real (and not imaginary as you'd find in Further Maths, so you cannot have f(i) as x=i is outside reals). x \in \mathbb{R} means x belongs to the set of real numbers. Since x is the domain, the domain is real.

    And if you get more error from sketching graphs, then stick to the methods I provided if you're okay with them.
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    (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 g(x)=\frac{1}{x} 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 g(x) for comparison:

    Name:  asdfasdfsadf.PNG
Views: 16
Size:  15.9 KB


    Your function of f(x)=\frac{2x+5}{x-3} is a similar to g(x) 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, x=3, as f(3) 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 f(x) as x tends to positive/negative infinity, therefore I'd suggest dividing (using long division or otherwise) 2x+5 by x-3 and expressing it in the form A+\frac{B}{x-3} which you'd find to be f(x)=2+\frac{11}{x-3} then consider what happens as x\rightarrow \pm \infty. 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 f(x) \rightarrow 2 as x\rightarrow \pm \infty .

    It never reaches 2, therefore it cannot be equal to it. Hence where the f(x) \in \mathbb{R}, f(x) \not= 2 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 x.

    You get left with \displaystyle f(x)=\frac{2+ \frac{5}{x} }{ 1- \frac{3}{x} }.

    Now if you take the limit as x \rightarrow \pm \infty then you can see that \frac{5}{x} AND \frac{3}{x} both go to 0, and your overall fraction is f(x) \rightarrow \frac{2}{1} \Rightarrow f(x) \rightarrow 2 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 \displaystyle h(x)=\frac{3x-5}{2x+1}
    >_< 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
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    (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
    Quite simple. You are finding a certain y coordinate that the graph approaches but never reaches as you keep going along the x-axis, so you are basically going off to infinity along the x-axis in order to get the asymptote. From our equation, we can see that as we go off to infinity, the denomintor will get bigger and bigger, so the overall fraction will go to 0.
 
 
 
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