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

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Answer 1 · 2017-02-07T12:11:55

See 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]}

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