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calculus

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Find the coordinates of all points on the graph of y=23−x2 at which the tangent line passes through the point (10,26).please show the work i greatly appreciate it :)

  • calculus -

    y = 23 - x2
    y' = -2x

    So, we want all points where the line through (x,y) and (10,26) has slope -2x

    (y-26)/(x-10) = -2x
    (23-x2 - 26)/(x-10) = -2x
    -3 - x2 = -2x(x-10)
    -3 - x2 = -2x2 + 20x
    x2 - 20x - 3 = 0
    x = -0.149 or 20.149

    At those values for x, we have
    (x,y) = (-0.149,22.978) and (20.149,-382.982)

    y' = 0.290 and -40.298

    the line through (-0.149,22.978) and (10,26) has slope 3.022/10.149 = 0.29

    the line through (20.149,-382.982) and (10,26) has slope 408.982/-10.149 = 40.298

    Look slike those are our lines.

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