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Math Help - Secant line

  1. #1
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    Secant line

    The point P(4,2) lies on the curve y = x^(1/2) . If Q is the point (x,x^(1/2)) , use your calculator to find the slope of the secant line PQ (correct to 6 decimal places) for the value x=3.99.

    How would I approach this?
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  2. #2
    Member Chokfull's Avatar
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    Take the formula for the slope between two points.


    <br /> <br />
m=(2-x^{\frac{1}{2}})/(4-x)<br /> <br />


    Then just input your value for x.
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    So it doesn't exist?
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    A Plied Mathematician
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    I think you'll find the slope does exist, and is not some huge number. 3.99 is not equal to 4, and you'll also see that the numerator is rather small as well. That's very typical of secant/tangent lines to smooth curves. You're always dealing with slopes that look like, or are close to, 0/0.
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  5. #5
    Member Chokfull's Avatar
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    Quote Originally Posted by bobsanchez View Post
    So it doesn't exist?
    Well, for it to not exist would require the square root of a negative number or 0 in the denominator. It may be 0 in the denominator if you meant .99 repeating, but this would still be an infinitesmally small number.
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  6. #6
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    Quote Originally Posted by bobsanchez View Post
    So it doesn't exist?
    you should get a value very close to 0.25
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