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    Sorry to be a pain again (afraid you'll have to get use to me pestering with questions while I attempt to teach myself some of P4-P6!).

    Could anyone tell me what the graph of y=(x^2 +10 - 7x)/x would look like? I found the roots of the equation and also the position of the assymptotes however from just looking at the equation I still wouldn't have a clue of the general shape of the curve unless I plotted a few points to help. Is there an easier way?

    Also, I'm trying to solve the following inequality:

    [(x+3)/(x-1)] < [(x-3)/(x+1)]

    and am trying to draw a graph of the curve of each of the separate sides to find the solution sets. However, I'm stuck as to how I go about drawing the curve. I know it's probably really easy but when I see (x+3)/(x-1) I don't have a clue about the shape! Is such a function the same shape as a 1/x graph and if so how do I work out the positioning of the assymptotes etc.

    Thanks for all your guys' help! I'm sure you don't really mind as it'll keep your brains ticking over the summer :rolleyes:
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    (Original post by Hoofbeat)
    Sorry to be a pain again (afraid you'll have to get use to me pestering with questions while I attempt to teach myself some of P4-P6!).

    Could anyone tell me what the graph of y=(x^2 +10 - 7x)/x would look like? I found the roots of the equation and also the position of the assymptotes however from just looking at the equation I still wouldn't have a clue of the general shape of the curve unless I plotted a few points to help. Is there an easier way?

    Also, I'm trying to solve the following inequality:

    [(x+3)/(x-1)] < [(x-3)/(x+1)]

    and am trying to draw a graph of the curve of each of the separate sides to find the solution sets. However, I'm stuck as to how I go about drawing the curve. I know it's probably really easy but when I see (x+3)/(x-1) I don't have a clue about the shape! Is such a function the same shape as a 1/x graph and if so how do I work out the positioning of the assymptotes etc.

    Thanks for all your guys' help! I'm sure you don't really mind as it'll keep your brains ticking over the summer :rolleyes:
    The best way to do the inequality is as follows:

    (x+3)/(x-1) = 1 + 4/(x-1),

    (x-3)/(x+1) = 1 - 4/(x+1),

    so you want 4/(x-1) < -4/(x+1), i.e 1/(x+1) + 1/(x-1) < 0, i.e 2x/(x+1)(x-1) < 0. Now try to solve...
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    y=(x^2 +10 - 7x)/x

    Factorise the top line into (x-2)(x-5)

    y = (x-2)(x-5)/x

    You now know that the graph will have an asymptote at 0, because y is undefined when you divide by 0. It will also be cross the x-axis at 2 and 5.

    As x tends to infinity y tends to infinity, and as x tends to -infinity y tends to -infinity.

    Some more ways to think about it include writing it out as

    y= x +10/x -7

    Now the -7 term is a constant and can be ignored for the point of sketching. We shall look at how x and 10/x look for small and big x. When x is small obviously the x term is small, but 10/x becomes large, so for small values of x i.e. |x|< 10 the graph will resemble y=10/x. But for larger x the graph resembles y=x.

    Equipped with this knowledge you should have enough to sketch it. I have included a graph drawn in mathematica to illustrate.
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    (Original post by Hoofbeat)
    ...
    and am trying to draw a graph of the curve of each of the separate sides to find the solution sets. However, I'm stuck as to how I go about drawing the curve. I know it's probably really easy but when I see (x+3)/(x-1) I don't have a clue about the shape! Is such a function the same shape as a 1/x graph and if so how do I work out the positioning of the assymptotes etc.
    ...:
    Ok, you can see here that there is a denominator that gets a zero value viz. x = 1.
    So the line x=1 is an asymptote.
    Looking at (x+3)/(x-1), then as x-> ∞, (x+3)/(x-1) -> x/x which = 1, i.e. y ->1.
    So the line y=1 is an asymtote.
    Rearranging the eqn,
    y = (x-1 + 4)/(x-1)
    y = 1 + 4/(x-1)
    Again you can see that as (x-1) -> ∞, y ->1.

    And, as you said, you can now see why it looks a bit like the 1/x function.
    You know that 1/x has asymptotes at x=0 and y=0, so just draw your function like 1/x, but using x=1 and y=1 as the asymptotes.
 
 
 
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