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

Feb 7, 2017

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.

$y ' = - 8 {x}^{3} + 9 {x}^{2} = {x}^{2} \left(9 - 8 x\right) = 0$, at $x = 0 \mathmr{and} \frac{9}{8}$.

$y ' ' = 6 x \left(3 x - 4\right) = 0 , a t x = 0 \mathmr{and} \frac{4}{3}$.

$y ' ' ' = 12 \left(3 x - 2\right) \ne 0$, at any point with $x \ne \frac{2}{3}$.

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

At $x = \frac{9}{8} , y = 1.068 , y ' = 0 \mathmr{and} y ' ' < 0$.

So, y = 1.068, nearly, 0 is a local maximum at $x = \frac{9}{8} = 1.125$.

As $x \to \pm \infty , y = - {x}^{4} \left(2 - \frac{1}{x}\right) \to - \infty$

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