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    The curve of f(x) = ax^2 + bx + c passes through the point (2, 24) and the gradient of the curve at this point is 22.
    The value of f''(x) is 6.
    Find the coordinates of the points where the curve crosses the x-axes and y-axes and of the minimum stationary point.
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    Put in 2 and 24. That gives you one equation for a,b,c

    Differentiate, put in 2 and 22, and that gives you an equation for a and B

    Differentiate again , put in 2 and 6 for the values and a falls straight out.

    Once you have a,b and c, it is easy to find the intercepts and the stationary point.
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    f(2): 

\Rightarrow 24=4a+2b+c

    f'(x)=2ax+b
    f'(2):

\Rightarrow 22=4a+b

    f''(x)=2a
    \Rightarrow 6=2a
    \therefore a=3

    22=12+b
    \therefore b=10

    24=12+20+c
    \therefore c=-8

    Hence
    f(x)=3x^2+10x-8

    Y-Int: Let x=0
    f(0)=-8
    Therefore, y-int is (0,-8)

    X-Int: When f(x)=0
    0=(x+4)(2x-3)
    x=-4orx=\frac{2}{3}
    Therefore, x-ints occur at (-4,0) and (\frac{2}{3},0)

    f'(x)=6x+10
    For stationary pts, let f'(x)=0
    0=2(3x+5)
    x=\frac{-5}{3}

    f(\frac{-5}{3})=\frac{-49}{3}
    Therefore, stationary point occurs at (\frac{-5}{3}),\frac{-49}{3})

    Hehehe. I was bored so i decided to post this up. =]
 
 
 
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Updated: August 9, 2009

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