How do you sketch the curve $y=x^3+6x^2+9x$ by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

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Answer 1 · 2017-01-18T03:57:28

See graph and explanation

Polynomial graphs have no asymptotes.

As x to +-oo, y = x^3(1+3/x)^2 to +-oo, showing end behavior of

$uarr and darr$ , without limit.

So, there are no global extrema.

$y=x(x+3)^2=0$ , at $x = 0 and -3$ .

x-intercepts: $0 and -3$ .

$y'=3(x+1)(x+3)=0$ , at $x = -3 and -1$

Turning points or points of inflexion at (-1, -4) and (-3, 0)

$y''=6x+12=0$ , at $x = -2$ .

$y'''=6 ne 0$ .

At $(-2, -2), y''=0 and y''' ne 0$ . So, it is the point of inflexion.

This POI is marked, in the graph.

At $(-1, -4)), y'=0 and y''=6> 0$ . So, local minimum $y = -4$ .

At $(-3, 0), y'=0 and y''= -6$ . So, $0$ is a local maximum.

graph{(x(x+3)^2-y)((x+2)^2+(y+2)^2-.01)=0 [-10, 10, -10, 5]}

How do you sketch the curve y=x^3+6x^2+9xy=x^3+6x^2+9x by finding local maximum, minimum, inflection points, asymptotes, and intercepts?