# How do you graph by using the zeros for g(x)=x^3-3x^2?

Jun 22, 2017

See below

#### Explanation:

$g \left(x\right) = {x}^{3} - 3 {x}^{2}$

Firstly we are asked to find the zeros of $g \left(x\right)$

$g \left(x\right) = 0 \to {x}^{3} - 3 {x}^{2} = 0$

${x}^{2} \left(x - 3\right) = 0$

$\therefore x = 0 \mathmr{and} x = 3$ are the zeros of $g \left(x\right)$

To graph $g \left(x\right)$ it now becomes necessary to find the turning point(s). Since this question is in "Precalculus" section we will observe the graph of $g \left(x\right)$ (below) rather than analysing the function using calculus.

graph{x^3-3x^2 [-10, 10, -5, 5]}

From the graph, we can see the zeros at $x = 0$ and $x = 3$ and the local minimum at $\left(2 , - 4\right)$

Jun 22, 2017

$\text{see explanation}$

#### Explanation:

$\text{ set " g(x)=0" for zeros}$

${x}^{3} - 3 {x}^{2} = 0 \Rightarrow {x}^{2} \left(x - 3\right) = 0$

$x = 0 \text{ multiplicity 2}$

$x = 3 \text{ multiplicity 1}$

$\text{we require more than the zeros to draw the graph}$

$\text{differentiate " g(x)" for stationary points and nature}$

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

$\text{equate to zero for stationary points}$

$3 {x}^{2} - 6 x = 0 \Rightarrow 3 x \left(x - 2\right) = 0$

$\text{stationary points at " x=0" and } x = 2$

$g \left(0\right) = 0 \Rightarrow \left(0.0\right) \text{ is a stationary point}$

$g \left(2\right) = 8 - 12 = - 4 \Rightarrow \left(2 , - 4\right) \text{ is a stationary point}$

$\text{using "color(red)"second derivative test}$

$g ' ' \left(x\right) = 6 x - 6$

$g ' ' \left(0\right) = - 6 \Rightarrow \left(0 , 0\right) \textcolor{red}{\text{ is a maximum}}$

$g ' ' \left(2\right) = 6 \Rightarrow \left(2 , - 4\right) \textcolor{red}{\text{ is a minimum}}$

$\text{since degree of " g(x)" is odd and the leading }$
$\text{coefficient is positive}$

$\text{then graph starts down and ends up}$

$\text{combining the above gives the graph of } g \left(x\right)$
graph{x^3-3x^2 [-10, 10, -5, 5]}