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

Jan 18, 2017

See graph and explanation

#### Explanation:

Polynomial graphs have no asymptotes.

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

$\uparrow \mathmr{and} \downarrow$, without limit.

So, there are no global extrema.

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

x-intercepts: $0 \mathmr{and} - 3$.

$y ' = 3 \left(x + 1\right) \left(x + 3\right) = 0$, at $x = - 3 \mathmr{and} - 1$

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

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

$y ' ' ' = 6 \ne 0$.

At $\left(- 2 , - 2\right) , y ' ' = 0 \mathmr{and} y ' ' ' \ne 0$. 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 $\left(- 3 , 0\right) , y ' = 0 \mathmr{and} y ' ' = - 6$. So,$0$ is a local maximum.

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