# Thread: Consider the cubic y = f(x) = x^2(ax-b)

1. ## Consider the cubic y = f(x) = x^2(ax-b)

Consider the cubic y = f(x) = x^2(ax-b) where a > 0 and b < 0
Sketch the graph of y = f(x)
How would I do this?

Find the coordinates of the local maximum, local minimum and inflection points in terms of a and b.
I know that the first and second derivative needs to be calculated, then they are found when f'(x) and f''(x) equal 0
Any help given will be greatly appreciated
2. $f(x) = ax^3-bx^2$
$f'(x) = 3ax^2-2bx$.
$f'(x)=0$ gives $x = 0$ or $x=\frac{2b}{3a}$ which is known to be negative since $a>0$ and $b<0$.
If you look at the sign of f'(x), you see that it is positive before $\frac{2b}{3a}$, negative between $\frac{2b}{3a}$ and 0, and then positive again after zero. This comes from the fact that the coefficient of $x^2$ is positive This implies that the function is increasing, decrease and then increasing. To sketch your graph, compute f at zero and at $\frac{2b}{3a}$ and then mark the points you got as local maximum and local minimum. Now draw the graph following the variations of f.
For the inflection point, $f''(x)=6ax-2b$. Setting this to zero gives $x=\frac{b}{3a}$. Plugging this into f gives the coordinates of the inflection point.