The normal to the curve y = ax^1/2 + bx at the point where x = 1 has a gradient 1 and intercepts the y-axis (0,-14). Find a and b

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  1. y = a√x + bx
    y' = a/(2√x) + b

    Since the normal at x=1 has slope 1, the tangent has slope -1. So,

    y'(1) = a/2 + b = -1
    b = -(a+2)/2

    y(1) = a - (a+2)/2 = (a-2)/2

    The normal line through (1,(a-2)/2) is thus

    y-(a-2)/2 = 1(x-1)
    This passes through (0,-14), so
    -14 - (a-2)/2 = 1(-1)
    a = -24
    So, b = 11

    y = -24/√x + 11x
    y(1) = -13
    the normal at (1,-13) is
    y+13 = 1(x-1)
    y = x-14
    This passes through (0,-14)

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