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find a function y=ax^2+bx+c whose graph has an x-intercept of 1, a y-intercept of -2, and a tangent line with slope -1 at the y-intercept.

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  1. The three given conditions are:

    1. For x-intercept of 1:
    f(x)=y= (x-1)(ax-c) = ax²-(c+1)+c
    2. For y-intercept of -2:
    c=-2
    3. for tangent=-1 at y=0
    Now find
    f'(x)=2ax+b
    to satisfy
    f'(0)=-1, or
    2a(0)+b=-1

    Solve for a, b and c.

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