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Math Help - Extrema, Intercepts, Symmetry, Asymptotes

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    Extrema, Intercepts, Symmetry, Asymptotes

    Sketch the graph of the equation. Look for extrema, intercepts, symmetry, and asymptotes as necessary.

    y=(2+x)/(1-x)

    Please Help
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  2. #2
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    Quote Originally Posted by rawkstar View Post
    Sketch the graph of the equation. Look for extrema, intercepts, symmetry, and asymptotes as necessary.

    y=(2+x)/(1-x)

    Please Help
    all of this part is precalculus stuff ...

    y-intercept is (0,2)

    x-intercept is (-2,0)

    vertical asymptote is x = 1

    horizontal asymptote is y = -2


    calculus part ...

    y' = \frac{3}{(1-x)^2} > 0 for all x \ne 1

    what's that tell you?

    y'' = \frac{6}{(1-x)^3}

    y''(x) < 0 for x > 1

    y''(x) > 0 for x < 1

    what does that tell you?
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    it's a simple one-variable function to investigate...

    you can open it to y=-x^2-x+2

    and then derivate it y=-2x-1.
    comparing the derivative to 0 will find extrema x=-1/2
    and since the function is well defined for all x, this is the only
    extrema.

    intercept with x axis is in the two roots which are conviniently
    shown to you x=-2, x=1
    and with y axis is simply y=2 (put 0 instead of x)

    it is symmetrical, and the symmetry axis is the straight line x=-1/2, since it is a squre function.

    and there are no asymptotes here, since the function is well defined
    for all x, and no asymptote as x--->infinity because it is a square function.

    that's all, you need to focus your question to the part you're having
    trouble with, because i'm not sure which one i needed to answer,
    and if, at all, i've answered your question.
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  4. #4
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    Quote Originally Posted by vonflex1 View Post
    it's a simple one-variable function to investigate...

    you can open it to y=-x^2-x+2
    uhh ... the function is y = \frac{2+x}{1-x} , not y = (2+x)(1-x)
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    Quote Originally Posted by skeeter View Post
    all of this part is precalculus stuff ...

    y-intercept is (0,2)

    x-intercept is (-2,0)

    vertical asymptote is x = 1

    horizontal asymptote is y = -2


    calculus part ...

    y' = \frac{3}{(1-x)^2} > 0 for all x \ne 1

    what's that tell you?

    y'' = \frac{6}{(1-x)^3}

    y''(x) < 0 for x > 1

    y''(x) > 0 for x < 1

    what does that tell you?
    I'm not sure how you found the horizontal asymptote to be -2, i got -1 by dividing everything by x
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  6. #6
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    Quote Originally Posted by rawkstar View Post
    I'm not sure how you found the horizontal asymptote to be -2, i got -1 by dividing everything by x
    you're right ... I was staring at the 2 in the numerator.
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