# 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
Thanks in advance

2. $\displaystyle f(x) = ax^3-bx^2$
$\displaystyle f'(x) = 3ax^2-2bx$.
$\displaystyle f'(x)=0$ gives $\displaystyle x = 0$ or $\displaystyle x=\frac{2b}{3a}$ which is known to be negative since $\displaystyle a>0$ and $\displaystyle b<0$.
If you look at the sign of f'(x), you see that it is positive before $\displaystyle \frac{2b}{3a}$, negative between $\displaystyle \frac{2b}{3a}$ and 0, and then positive again after zero. This comes from the fact that the coefficient of $\displaystyle x^2$ is positive This implies that the function is increasing, decrease and then increasing. To sketch your graph, compute f at zero and at $\displaystyle \frac{2b}{3a}$ 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, $\displaystyle f''(x)=6ax-2b$. Setting this to zero gives $\displaystyle x=\frac{b}{3a}$. Plugging this into f gives the coordinates of the inflection point.
hope this helps