Question

How do you use end behavior, zeros, y intercepts to sketch the graph of `g(x)=-x^3(2x-3)`?

Answer 1

See explanation.

Explanation:

Data for sketching :
y-intercept : 0

x-intercept : 0 and 3/2

So, A(0, 0) and B(3/2, 0) lie on the graph.

`y'=-8x^3+9x^2=x^2(9-8x)=0`, at `x = 0 and 9/8`.

`y''=6x(3x-4)=0, at x = 0 and 4/3`.

`y'''=12(3x-2) ne 0`, at any point with `x ne 2/3`.

So, origin (0, 0) and (4/3, 64/81) are POI (points of inflexion ).

At `x = 9/8, y = 1.068, y' = 0 and y''<0`.

So, y = 1.068, nearly, 0 is a local maximum at `x =9/8=1.125`.

As `x to +- oo, y =-x^4(2-1/x) to -oo`

graph{(3x^3-2x^4-y)=0 [-5, 5 -2.5, 2.5]}