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Math Help - Slant asymptote

  1. #1
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    Slant asymptote

    x^n - n^2
    x^n-1 - n

    Does this statement have a slant asymptote for all values of n, if n is a natural number



    Everything in bold is in superscript
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  2. #2
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    Quote Originally Posted by shane99 View Post
    x^n - n^2
    x^n-1 - n

    Does this statement have a slant asymptote for all values of n, if n is a natural number



    Everything in bold is in superscript
    I presum you mean \frac{x^n- n^2}{x^{n-1}- n}. Just go ahead and divide it out: x^{n}- n^2 divided by x^{n-1}-n is equal to x+ a fraction with lower degree in the numerator than in the denominator. Thus, y= x+ n is a slant asymptote.

    The only "technical" points are for n= 2, where the fraction is \frac{x^2- 4}{x- 2} which is equal to x+ 2 for all x not equal to 2- would you call that an asymptote?- and n= 1 where the fraction is \frac{x- 1}{1- 1} which does not exist.
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  3. #3
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    Quote Originally Posted by shane99 View Post
    x^n - n^2
    x^n-1 - n

    Does this statement have a slant asymptote for all values of n, if n is a natural number



    Everything in bold is in superscript
    1. Rewrite the term of the function:

    f_n(x)=\dfrac{x^n-n^2}{x^{n-1}-n}=\dfrac{x^n-n^2}{\frac1x \cdot x^n-n} = \dfrac{x^{n+1}-n^2 x}{x^n - nx}

    2. Do long division:

    f_n(x)= \dfrac{x^{n+1}-n^2 x}{x^n - nx} = x-\dfrac{n^2 x}{x^n-nx} + \dfrac{nx^2}{x^n-nx}

    3. f_1(x) doesn't exist for n = 1 because the denominator would be zero.

    f_2 (x) (that means n = 2) has the asymptote y = x+2

    For all n > 2 the fractions in the term become zero if x approaches infinity. Therefore the asymptote is y = x.


    EDIT: Beaten! Again
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