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Math Help - area of intersecting curves

  1. #1
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    area of intersecting curves

    some help with this question would be much appreciated.

    A curve y = f (x) has derivative f'(x) = 2x + 3 and intersects the line 5y - 18x - 64 = 0 at the points (2,20) and (-3,10). Find the area enclosed by the curve f(x) and the line.
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  2. #2
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    Quote Originally Posted by 24680 View Post
    some help with this question would be much appreciated.

    A curve y = f (x) has derivative f'(x) = 2x + 3 and intersects the line 5y - 18x - 64 = 0 at the points (2,20) and (-3,10). Find the area enclosed by the curve f(x) and the line.
    If y'=2x+3 then,
    y=x^2+3x+C, for some C ----> Anti-derivative

    The problem says the curve (parabola in this case) intersects at (2,20) which means when x=2 then y=20:
    20=4+6+C thus, C=10.
    The unique curve must be,
    y=x^2+3x+10
    The line is,
    y=3.6x+12.8
    The problem is that these two curve intersect in one other point which is not (-3,10) as you say. Thus there is no such curve.
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  3. #3
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    Hello, 24680!

    There are typos in this problem. .As written, it is contradictory.
    Please check the original wording.


    A curve y = f(x) has derivative f '(x) = 2x + 3
    and intersects the line 5y - 18x - 64 = 0 at the points (2,20) and (-3,10).
    Find the area enclosed by the curve f(x) and the line.

    Contradiction . The point (-3,10) is not on the line 5y - 18x - 64 = 0


    I suspect that you gave the wrong equation for the line.

    The line through (2,20) and (-3,10) is: .2x - y + 16 .= .0

    The curve turns out to be the parabola: .y .= .x + 3x + 10

    Am I close?

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