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    The parabola y= ax^2 + bx + c passes through point (1,1) and has a tangent with gradient 7 at point (3,3).

    Find the values of a, b and c.

    Please help?

    So far: I've done

    f(x) = ax^2 + bx + c
    f'(x) = 2ax + b

    Bit stuck on what to substitute into those equations above.

    Thank you!
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    y = ax^2 + bx + c passes through the point (1,1), so when x = 1, y = 1. It also passes through (3,3), so when x = 3, y = 3. Subtract the two resultant equations to get an equation in a and b. Then y' = 2ax + b, and when x = 3, the gradient of the tangent is 7, so y' = 7 - this gives you another equation for a and b. Solve simultaneously.
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    (Original post by generalebriety)
    y = ax^2 + bx + c passes through the point (1,1), so when x = 1, y = 1. It also passes through (3,3), so when x = 3, y = 3. Subtract the two resultant equations to get an equation in a and b. Then y' = 2ax + b, and when x = 3, the gradient of the tangent is 7, so y' = 7 - this gives you another equation for a and b. Solve simultaneously.
    You're brilliant! Thank you for your help! :yep:
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    No problem.
 
 
 
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Updated: February 2, 2010
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