C1 Sketching Curves HELP

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    How do I start question 3? Thanks
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    g(x)=f(x-3) = (x-3)/(x-4), so the asymptotes are 1 and 4 (this is a translation by 3 units in the x-direction)

    For b, we solve the equation (x-3)/(x-4)= 0 to give x=3 as x -intercept. and find g(0) = y= 3/4 for y-intercept
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    (Original post by 123Master321)
    g(x)=f(x-3) = (x-3)/(x-4), so the asymptotes are 1 and 4 (this is a translation by 3 units in the x-direction)

    For b, we solve the equation (x-3)/(x-4)= 0 to give x=3 as x -intercept. and find g(0) = y= 3/4 for y-intercept
    Oh thank you but the only bit I don't understand is how you found the horizontal asymptote
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    (Original post by Amarante)
    Oh thank you but the only bit I don't understand is how you found the horizontal asymptote
    You need to know that  \displaystyle \lim_{x\to \infty } \frac{x-3}{x-4} =1 .
    If you're not sure about this then divide numerator and denominator by x to get  \displaystyle \frac{1-\frac{3}{x}}{1-\frac{4}{x}} .
    You know that  1/x \rightarrow \infty as  x \rightarrow \infty so the result follows.
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    I think it's something to do with y=f(x), and if x doesn't equal 0, can we assume that the asymptote for x is the same as y? Then the horizontal asymptote would be y = 1.
    I'm just in C1 as well, so I'm not really sure about this off the top of my head. If you know someone in Year 13 who takes Maths it might be a good idea to ask them.
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    to find the horizontal asymptote you let x get really really big ?

    ...enormous x -3 is really just x

    .... enormous x - 4 is really just x

    so dividing { x - 3 } / { x - 4 } is pretty much 1
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    (Original post by B_9710)
    You need to know that  \displaystyle \lim_{x\to \infty } \frac{x-3}{x-4} =1 .
    If you're not sure about this then divide numerator and denominator by x to get  \displaystyle \frac{1-\frac{3}{x}}{1-\frac{4}{x}} .
    You know that  1/x \rightarrow \infty as  x \rightarrow \infty so the result follows.
    I understand now, appreciate your help but this doesn't seem like a c1 topic
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    (Original post by Amarante)
    I understand now, appreciate your help but this doesn't seem like a c1 topic
    If you want to make it C1, you could just sketch the original graph and note the horizontal asymptote wont change after translation
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    You can divide to get  \displaystyle f(x)=1+ \frac{1}{x-1} .
    Consider now the 2 transformations that map the curve y=1/x to y=f(x).
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    (Original post by B_9710)
    You know that  1/x \rightarrow \infty as  x \rightarrow \infty so the result follows.
    1/x \rightarrow 0 as  x \rightarrow \infty

    (Original post by Amarante)
    I understand now, appreciate your help but this doesn't seem like a c1 topic
    It is a perfectly valid technique for finding the horizontal asymptote so you can use it. Finding the horizontal asymptote is just taking a limit, so any other method would be an adaptation of this.
 
 
 
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