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Math Help - How to find intercepts, asymptotes, end behaviors, etc.

  1. #1
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    How to find intercepts, asymptotes, end behaviors, etc.

    1) 2
    x^2-2x-3
    Find) Domain
    Range
    Is This Continuous?
    Local min/max
    Vertical Asymptote
    Horizontal Asymptote
    X Intercepts
    Y Intercepts
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  2. #2
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    for function f(x) = \frac{2}{x^2-2x-3}

    f(x) = \frac{2}{(x+1)(x-3)}

    Domain
    Domain is the set of values of x for which function is defined.
    The given function is not defined at x = -1 and x = 3

    So, domain = \{x \in \mathbb{R} | \;x\ne-1,3 \}

    Range
    Range is f(x) values (or y values) you get, when you put these domain x values in function f(x). Since the function has a horizontal Asymptote at y = 0, so, y = 0 is excluded from Range.

    So, Range = \{y \in \mathbb{R} | \;y\ne0\}

    Horizontal Asymptote (HA)

    To find HA, divide Numerator and denominator of the function with highest degree (power) of x. Then take limit x \rightarrow \infty

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

    divide with highest degree, x^2,

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

    f(x) = \frac{\frac{2}{x^2}}{1-\frac{2}{x}-\frac{3}{x^2}}

    Now take \lim{x\to \infty}

    = \frac{0}{1-0-0}=0

    HA is y = 0

    Vertical Asymptote (VA)

    For VA, Denominator = 0

    (x+1)(x-3) = 0

    x = -1 and x = 3

    The graph is discontinuous at x = -1 and x = 3

    y-intercepts

    For y-intercept, put, x = 0 in equation,

    we get, y = -2/3

    y-intercept = -2/3, the graph cuts y-axis at -2/3.

    x-intercepts

    For x-intercept, put, y = 0 in equation,

    0 = \frac{2}{(x+1)(x-3)}

    we do not get any answer from this, so there is no, x-intercept. i.e., the graph does not cuts x-axis

    Please see attached graph, the green lines show the asymptotes.
    Attached Thumbnails Attached Thumbnails How to find intercepts, asymptotes, end behaviors, etc.-graph12.jpg  
    Last edited by Shyam; October 25th 2008 at 02:39 PM.
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