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    hey, i dont really understand pt ii of the question. i was wondering how the graph appraoches the y asymptote for large values of x but idk what to do :/ the mark scheme is giving me confusing thoughts

    btw the graph is y= 3x^2/(2x-1)(x+2)
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    (Original post by penelopecrux)
    hey, i dont really understand pt ii of the question. i was wondering how the graph appraoches the y asymptote for large values of x but idk what to do :/ the mark scheme is giving me confusing thoughts

    btw the graph is y= 3x^2/(2x-1)(x+2)
    For very large values of x does the curve approach the asymptote from above or below? same for very small values of x.

    Just sub in a really big number and a really small number and you should be able to figure it out. This should also help you sketch the rest of the graph.
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    (Original post by Mikasadoge)
    For very large values of x does the curve approach the asymptote from above or below? same for very small values of x.

    Just sub in a really big number and a really small number and you should be able to figure it out. This should also help you sketch the rest of the graph.
    no im afraid that doesnt work
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    (Original post by penelopecrux)
    hey, i dont really understand pt ii of the question. i was wondering how the graph appraoches the y asymptote for large values of x but idk what to do :/ the mark scheme is giving me confusing thoughts

    btw the graph is y= 3x^2/(2x-1)(x+2)
    Investigate where your graph is around x=\frac{1}{2}. We know that for x \rightarrow \frac{1}{2}_- we have y \rightarrow - \infty on the graph. And for x \rightarrow \frac{1}{2}_+ we have x^2 > 0 and 2x-1 > 0 and x+2 > 0 so the overall fraction will be +ve hence y \rightarrow \infty. So, note that after x=1/2 the graph comes in from +ve infinity.
    Then, we must investigate whether it crosses the line y=3/2 or not. This is straightforward as you can set \dfrac{3}{2} = \dfrac{3x^2}{(2x-1)(x+2)} hence obtain 0=9x-6 which has an obvious solution, meaning the curve DOES intersect the line exactly once.
    The only way a branch can satisfy all of these is if it comes down from infinity, crosses it, then comes back up but only gets closer to the asymptote from below it.

    Hence, this branch is:

    Spoiler:
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