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Math Help - Extra-credit problem

  1. #1
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    Extra-credit problem

    can anyone help me in graphing this ?

    f(x) = whole sqrt x + 3/x - 2
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  2. #2
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    Quote Originally Posted by darkangel View Post
    can anyone help me in graphing this ?

    f(x) = whole sqrt x + 3/x - 2
    Is it f(x) = whole sqrt x + (3/x) - 2 or f(x) = whole sqrt x + 3/(x - 2)
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  3. #3
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    i am sorry about that
    its f(x) = whole sqrt (x + 3)/(x - 2)
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  4. #4
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    Quote Originally Posted by darkangel View Post
    i am sorry about that
    its f(x) = whole sqrt (x + 3)/(x - 2)
    y = \sqrt{\frac{x+3}{x-2}} = \sqrt{\frac{[x-2]+5}{x-2}} = \sqrt{1 + \frac{5}{x-2}}.

    First of all you should establish the implied domain. You need x + 3 \geq 0 AND x - 2 > 0 => ......, OR x + 3 \leq 0 AND x - 2 < 0 => ......

    There's no y-intercept (why?).

    x-int: Solve x + 3 = 0 (why?).

    Vertical asymptote is got from solving x - 2 = 0 (why?).

    Horizontal asymptote: As x \rightarrow \pm \infty, y --> 1 (why?). In fact, as x --> -oo, y --> 1 from below (why?) and as x --> +oo, y --> 1 from above (why?)

    Stationary points: Solve dy/dx = 0. (Hint: There are none).

    Now draw the shape that ties all these features together .....
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  5. #5
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    thank u so much for ur response...
    umm i did not understand about the horizontal asymtote ( like how can i find it, if i get one of these problems on the test )
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  6. #6
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    nevermind.. i figured it out
    thanks again for ur answer
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