# How do you sketch the graph y=x^3-4x^2 using the first and second derivatives?

Dec 20, 2016

$y = {x}^{3} - 4 {x}^{2}$

1. General Observations

A cubic function and the cubic coefficient is +ve, so it will have the classic cubic shape with a maximum to the left of a minimum,

2. Roots

$y = 0 \implies {x}^{3} - 4 {x}^{2} = 0$
$\therefore {x}^{2} \left(x - 4\right) = 0$
 :. x=0 "(double root"), x=4

So we can already deduce that there must be a local maximum at $x = 0$ that touches the axis (because of the double root) and a local minimum in between $x = 0$ and $x = 4$ (or else it could not be a cubic with a +ve cubic coefficient).

3. Turning Points

$y = {x}^{3} - 4 {x}^{2} \implies y ' = 3 {x}^{2} - 8 x$
At min/max $y ' \left(0\right) = 0 \implies 3 {x}^{2} - 8 x = 0$
$\therefore x \left(3 x - 8\right) = 0$
$\therefore x = 0 \text{(as expected)} , x = \frac{8}{3} \left(\approx 2.7\right)$

When $x = 0 \implies y = 0$
When $x = \frac{8}{3} \implies y = \frac{512}{27} - \frac{256}{9} = - \frac{245}{2} \left(\approx - 9.5\right)$

4. Nature of Turning Points

$y ' = 3 {x}^{2} - 8 x \implies y ' ' = 6 x - 8$

When $x = 0 \implies y ' ' < 0 \implies \text{maximum (as expected)}$
When $x = \frac{8}{3} \implies y ' ' > 0 \implies \text{minimum (as expected)}$

5. The Graph

There is now information to plot the graph, here I will use the actual graph:

graph{x^3-4x^2 [-10, 10, -15, 10]}