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Math Help - Finding intercepts

  1. #1
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    Finding intercepts

    ok so i need to find the intercepts, includes x and y, of this equation y=(x-1) times the square root of (x^2+1)
    so far when i try doing it i don't get the points which are shown in the calculator
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  2. #2
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    Quote Originally Posted by la_lo626 View Post
    ok so i need to find the intercepts, includes x and y, of this equation y=(x-1) times the square root of (x^2+1)
    so far when i try doing it i don't get the points which are shown in the calculator
    To find the y-intercept, set x=0.

    y=(0-1)\sqrt{0^2+1} = -1

    The y-intercept is at point (0,-1).

    To find the x-intercept, set y=0.

    0=(x-1)\sqrt{x^2+1}

    Since \displaystyle\sqrt{x^2+1} can't equal 0, we must have x = 1.

    The x-intercept is at point (1,0).
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  3. #3
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    Quote Originally Posted by undefined View Post
    To find the y-intercept, set x=0.

    y=(0-1)\sqrt{0^2+1} = -1

    The y-intercept is at point (0,-1).

    To find the x-intercept, set y=0.

    0=(x-1)\sqrt{x^2+1}

    Since \displaystyle\sqrt{x^2+1} can't equal 0, we must have x = 1.

    The x-intercept is at point (1,0).

    i am confused in the x-intercept part. the answer is cosrrect but i was wondering how did you get to that point i'm just not sure is there a step in which you're sppose to factor out the equation?
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  4. #4
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    Quote Originally Posted by la_lo626 View Post
    i am confused in the x-intercept part. the answer is cosrrect but i was wondering how did you get to that point i'm just not sure is there a step in which you're sppose to factor out the equation?
    Suppose "a" and "b" are real numbers and ab=0. Then either a=0 or b=0 (or both). Can you see why \displaystyle\sqrt{x^2+1} can't possibly be 0?

    So x-1 = 0 \implies x=1.
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  5. #5
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    thanks "undefined" i see now sort of it's a little more clear
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