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: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