Given a 3. degree function:
$f(x) = ax^3 + bx^2 + cx + d$
Furthermore, I am given
- the graph of function $f(x)$ passes through x-axis at origin.
- the tangent at point P(-3, 0) is parallel with line $y = 6x$
How can I solve for a, b, c and d ?
So far I have,
Since graph passes through origin, $f(0) = 0$ so, $d = 0$ ----------- (1)
Graph also passes through P(-3, 0) so I get,
$0 = -27a + 9b -3c + d$ ----------- (2)
Now to get tangent of line at P(-3, 0) taking first derivative w.r.t $x$ I get,
$f'(x) = 3ax^2 + 2bx +c$
Since tangent is parallel to $y = 6x$, we get slope $m = 6$
$6 = 27a - 6b + c$ ----------- (3)
How can I get fourth equation to solve for all the values.
EDIT 2: Condition 1 (first bullet point), in German:
- Die Graph von f berührt die x-Achse im Ursprung
EDIT: Could the catch be in the first condition where it explicitly mentions that curve passes the 'x-axis' at origin. Giving another condition for curvature/maxima of the curve. Just my assumption here.