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Finding the equation of a line

  1. Feb 18, 2007 #1
    Hi there,

    I have to complete the following question, but I have no idea how to approach it (there are four other parts to the question that I can't complete until I get the first bit). I have made numerous attempts, but am not sure how to get the a part of the equation.

    All help would be appreciated!


    A curve f(x) is defined by the equation : y = ax² + bx + c, where a, b and c are constants.

    The curve crosses the y-axis at the point (0,4). At this point the gradient of the graph is -5.

    The curve crosses the x-axis at point (1,0).

    (i) Find the values of a, b, and c and write down the equation of the curve


    Sub point (0.4) into equation to get c (x=0,y=4):

    y = ax² + bx + c
    4 = 0 + 0 + c
    c = 4

    If gradient at point (0,4) is -5, then dy/dx must be equal to -5.

    dy/dx = 2ax + b
    -5 = 2ax + b
    -5 = 2a(0) + b
    -5 = 0 + b
    b = -5

    *not sure about the bit below*

    c=-5, b=4, so sub these into equation of curve and use a point to find a

    y = ax² + bx + c

    at (1,0) x=1, y=0

    y = ax² + bx + c
    0 = a1² + (-5 x 1) + 4
    0 = a -5 + 4
    0 = a - 1
    a = 1 ???
  2. jcsd
  3. Feb 18, 2007 #2


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    Looks good to me :approve:
  4. Feb 18, 2007 #3


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    Now if only PHYSICS students will recognize this as a projectile problem...
  5. Feb 19, 2007 #4


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    Kafka, you did that exactly right.

    It does, however, have nothing to do with "differential equations" so I am moving it to the Calculus section.
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