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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- May 26th 2006, 08:28 PMicarus[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? - May 27th 2006, 12:51 AMCaptainBlackQuote:

Originally Posted by**icarus**

$\displaystyle

y=ax^3+bx^2+cx+d

$

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

$\displaystyle

a+b+c+d=1\ \ \ \dots(1)

$

The derivative of the cubic is:

$\displaystyle

\frac{dy}{dx}=3ax^2+2bx+c

$,

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

$\displaystyle

3a+2b+c=2\ \ \ \dots(2)

$.

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

the equation:

$\displaystyle

\frac{dy}{dx}=3ax^2+2bx+c=0

$

has a single root. For the quadratic to have a single root the discriminant

must be zero, therefore this condition implies that:

$\displaystyle

(2b)^2-4(3a)c=0

$

or:

$\displaystyle

b^2=3ac\ \ \ \dots(3)

$.

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