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Math Help - asymptote

  1. #1
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    asymptote

    ok so i've been given some isotopes and i'm not sure how to plot them, well i do but i'm kinda confuse .

    ok so i've ben given this h(x) = 1/2x + 1
    but the main one i'm confuse on its this -3 / 2x - 5 (totally dont get it)

    like i know for the first one the vertical asotopes is -1/2 and the second one is 5/2. Cause i know that most of the time u have to draw the isotopes base on a given point. So how do u find that point? thanks!
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by lickman View Post
    ok so i've been given some isotopes and i'm not sure how to plot them, well i do but i'm kinda confuse .

    ok so i've ben given this h(x) = 1/2x + 1
    but the main one i'm confuse on its this -3 / 2x - 5 (totally dont get it)

    like i know for the first one the vertical asotopes is -1/2 and the second one is 5/2. Cause i know that most of the time u have to draw the isotopes base on a given point. So how do u find that point? thanks!
    I believe you're talking about "asymptotes"

    h(x)=\frac{1}{2x+1}


    Look at the denominator. The denominator cannot be zero. So what would make the denominator zero?

    2x+1=0
    x=-\frac{1}{2}

    So, we have a vertical asymptote at x=-\frac{1}{2}

    In the other one, look at the denominator

    f(x)=\frac{-3}{2x-5}

    2x-5=0
    x=\frac{5}{2}

    Therefore, you have a vertical asymptote at x=\frac{5}{2}

    Just graph the vertical lines through these points.
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  3. #3
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    I think you mean h(x) = 1/(2x + 1)

    An asymptote exists where the graph is undefined. The graph approaches but never reaches this line.

    The graph is undefined when the denominator = 0 because we cannot divide by 0

    so asymptote exists when 2x +1 = 0 : 2x = -1 : x = -1/2

    in the second case an asymptote exists when 2x-5 = 0 : 2x = 5 : x=5/2
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  4. #4
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    i get what u mean , cause we're learning about reciprocal of a linear function, i know how graph the vertical asymotopes, but the thing i dont get is the curve for a function like this y = 1/x, rite. cause i thought most of the time theres a given point, and then u just draw ur curve base it on it. So the problem is i dunno how to draw the curve, correctly. thanks
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by lickman View Post
    i get what u mean , cause we're learning about reciprocal of a linear function, i know how graph the vertical asymotopes, but the thing i dont get is the curve for a function like this y = 1/x, rite. cause i thought most of the time theres a given point, and then u just draw ur curve base it on it. So the problem is i dunno how to draw the curve, correctly. thanks
    Assign arbitrary values for x (except 0, of course) and use the function to solve for y. Then plot the points.

    If x=1, then y=1
    If x=2, then y= 1/2
    If x=-1, then y=-1
    .
    .
    .
    etc.
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  6. #6
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    When graphing you may choose values of x eg. x=1 or x = 2 and find their corresponding y value

    if h(x) = 1/(2x+1)

    then h(1) = 1/(2(1)+1) = 1/3

    Note that for large x that the denominator becomes so large that h(x) approaches zero


    For very small x then x approaches zero and h(x) above approaches 1
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