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Math Help - [SOLVED] determining cubic rule

  1. #1
    icarus
    Guest

    [SOLVED] determining cubic rule

    Determine the family of cubics with
    the following characteristics: pass through the point (1,1), have
    a gradient of 2 when x = 1, and have 1 stationary point?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by icarus
    Determine the family of cubics with
    the following characteristics: pass through the point (1,1), have
    a gradient of 2 when x = 1, and have 1 stationary point?
    Let the cubic function be:

    <br />
y=ax^3+bx^2+cx+d<br />

    Then the first condition that the cubic pass through (1,1) means:

    <br />
a+b+c+d=1\ \ \ \dots(1)<br />

    The derivative of the cubic is:

    <br />
\frac{dy}{dx}=3ax^2+2bx+c<br />
,

    so the second condition the gradient is 2 when x=1 means:

    <br />
3a+2b+c=2\ \ \ \dots(2)<br />
.

    The third condition, that there be but one stationary point means that
    the equation:

    <br />
\frac{dy}{dx}=3ax^2+2bx+c=0<br />

    has a single root. For the quadratic to have a single root the discriminant
    must be zero, therefore this condition implies that:

    <br />
(2b)^2-4(3a)c=0<br />

    or:

    <br />
b^2=3ac\ \ \ \dots(3)<br />
.

    Now we have three equations in four unknowns, which usually means we
    can parameterise the set of solutions by one of the unknowns and solve
    for the others in terms of that unknown. Which is what you are required to do
    here.

    RonL
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