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Math Help - asymptotes

  1. #1
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    asymptotes

    hi guys.. need help in sketching graph and labeling intercepts and asymptotes of the attached function.

    it has a power 3 x in the denominator .. how can i solve and simplify it?

    thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by samtrix View Post
    hi guys.. need help in sketching graph and labeling intercepts and asymptotes of the attached function.

    it has a power 3 x in the denominator .. how can i solve and simplify it?

    thanks!
    y = \frac{x^3 + 4x^2 + 5x + 2}{x^3 -x^2 - 8x + 12}

    We may use the rational root theorem: The rational zeros of a polynomial
    ax^n + bx^{n - 1} + ~ ... ~ + cx + d
    (if they exist) will be of the form:
    \pm \frac{\text{factor of d}}{\text{factor of a}}

    So, for example, the possible rational zeros of the numerator are \pm 1, \pm 2. I find that both x = -1 and x = -2 are zeros of the polynomial. Thus
    x^3 + 4x^2 + 5x + 2 = (x + 1)(x + 2)(x + p)
    We can find the remaining zero by long division, synthetic division, or by multiplying the above factorized expression out and solving for p. I get:
    x^3 + 4x^2 + 5x + 2 = (x + 1)^2(x + 2)

    You can do a similar factorization of the denominator.

    -Dan
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  3. #3
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    Hello, samtrix!

    You should be familiar with the Factor Theorem and Remainder Theorem.
    (If you're not, I'll let someone else explain them.)


    Sketch the graph and label intercepts and asymptotes.

    y \:=\:\frac{x^3+4x^2 + 5x + 2}{x^3 - x^2 - 8x + 12}

    y-intercept: Let x = 0 and solve for y.

    .  y \:=\:\frac{0^3 + 4\!\cdot\!0^2+5\!\cdot\!9 + 2}{0^3-0^2-8\!\cdot\!0 + 12} \:=\:\frac{2}{12} \:=\:\frac{1}{6}\quad\Rightarrow\quad\boxed{ \text{y-intercept: }\left(0,\,\frac{1}{6}\right)}


    x-intercepts: Let y = 0 and solve for x.

    . . x^3 + 4x^2 + 5x + 2\:=\:0\quad\Rightarrow\quad(x+1)^2(x+2) \:=\:0

    Therefore: . x \:=\:-1,\,-2\quad\Rightarrow\quad\boxed{\text{x-intercepts: }(-1,\,0),\;(-2,\,0)}


    Vertical asymptotes: Value for which the denominator equals zero.

    . . x^3-x^2-8x + 12\:=\:0\quad\Rightarrow\quad(x-2)^2(x+3)\:=\:0

    Therefore, \boxed{\text{vertical asymptotes: }x \,=\,2,\;x\,=\,-3}


    Horizontal asymptote: Find \lim_{x\to\infty}y

    We have: . \lim_{x\to\infty}\frac{x^3+4x^2+5x+2}{x^3-x^2-8x+12}

    Divide top and bottom by x^3
    . . \lim_{x\to\infty}\frac{\frac{x^3}{x^3} + \frac{4x^2}{x^3} + \frac{5x}{x^3} + \frac{2}{x^3}}{\frac{x^3}{x^3} - \frac{x^2}{x^3} - \frac{8x}{x^3} + \frac{12}{x^3}} \;=\;\lim_{x\to\infty}\frac{1 + \frac{4}{x} + \frac{5}{x^2} + \frac{2}{x^3}}{1 - \frac{1}{x} - \frac{8}{x^2} + \frac{12}{x}} . =\;\frac{1 + 0 + 0 + 0}{1 - 0 - 0 + 0} \;=\;1

    Therefore: . \boxed{\text{Horizontal asymptote: }y \:=\:1}

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  4. #4
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    thanks soroban and topsquark! i got the idea ^__^

    and i will look up on factor theorem and remainder theoram.
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