# How do you sketch the graph of f(x)=x^3-3x^2?

Feb 24, 2015

Calculate derivative :

$f ' \left(x\right) = 3 {x}^{2} - 6 x = 3 x \left(x - 2\right)$

So you can study the sign of $f ' \left(x\right)$ :

• f'(x) < 0 iff x in ]0,2[
• f'(x) > 0 iff x in ]-oo, 0[ cup ]2,+oo[.

You get now the variations of $f$ :
- $f$ is decreasing on $\left[0 , 2\right]$
- $f$ is creasing on $\left[- \infty , 0\right]$ and on $\left[2 , + \infty\right]$.

Remark that $f$ has :
- local maximum in 0 with $f \left(0\right) = 0$
- local minimum in 2 with $f \left(2\right) = - 4$.
There are two horizontal tangents at 0 and at 2.

• For the limits, apply the rule :
${\lim}_{x \to - \infty} {x}^{3} - 3 {x}^{2} = {\lim}_{x \to - \infty} {x}^{3} = - \infty$
${\lim}_{x \to + \infty} {x}^{3} - 3 {x}^{2} = {\lim}_{x \to + \infty} {x}^{3} = + \infty$

Finally :

graph{x^3 - 3x^2 [-8.42, 13.78, -6.62, 4.48]}

Feb 24, 2015

You can start by setting $x = 0$ that gives you $y = f \left(0\right) = 0$ so your curve passes through the origin.
Setting $y = 0$ you get ${x}^{3} - 3 {x}^{2} = 0$ that gives $x = 0$ and $x = 3$
When $x \to + \infty$ $f \left(x\right) \to + \infty$ as well while when $x \to - \infty$ then $f \left(x\right) \to - \infty$.
Points of maximum or minimum are found by setting the first derivative equal to zero:
$f ' \left(x\right) = 3 {x}^{2} - 6 x$
and
$3 {x}^{2} - 6 x = 0$ gives: $3 x \left(x - 2\right) = 0$
and $x = 0$ and $x = 2$
When $x = 0$ $y = 0$ and when $x = 2$ $y = - 4$
Setting the second derivative equal to zero will give us inflection point(s):
$f ' ' \left(x\right) = 6 x - 6$
and
$6 x - 6 = 0$
so that $x = 1$ and $y = - 2$
And finally: