Question

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

Answer 1

See below

Explanation:

`g(x) = x^3-3x^2`

Firstly we are asked to find the zeros of `g(x)`

`g(x) = 0 -> x^3-3x^2=0`

`x^2(x-3)= 0`

`:. x =0 or x= 3` are the zeros of `g(x)`

To graph `g(x)` it now becomes necessary to find the turning point(s). Since this question is in "Precalculus" section we will observe the graph of `g(x)` (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 `(2, -4)`

Answer 2

`"see explanation"`

Explanation:

`" set " g(x)=0" for zeros"`

`x^3-3x^2=0rArrx^2(x-3)=0`

`x=0" multiplicity 2"`

`x=3" multiplicity 1"`

`"we require more than the zeros to draw the graph"`

`"differentiate " g(x)" for stationary points and nature"`

`g'(x)=3x^2-6x`

`"equate to zero for stationary points"`

`3x^2-6x=0rArr3x(x-2)=0`

`"stationary points at " x=0" and " x=2`

`g(0)=0rArr(0.0)" is a stationary point"`

`g(2)=8-12=-4rArr(2,-4)" is a stationary point"`

`"using "color(red)"second derivative test"`

`g''(x)=6x-6`

`g''(0)=-6rArr(0,0)color(red)" is a maximum"`

`g''(2)=6rArr(2,-4)color(red)" is a minimum"`

`"since degree of " g(x)" is odd and the leading "`
`"coefficient is positive"`

`"then graph starts down and ends up"`

`"combining the above gives the graph of " g(x)`
graph{x^3-3x^2 [-10, 10, -5, 5]}