# Identifying inflection points and critical points

I'm given an equation and asked to find the following:

• Domain
• Critical points
• Inflection points
• Asymptotes

The function is:

$y = 2 + 9x + 3x^2 -x^3$

It has been quite awhile since I have done this sort of problem so I'm a bit rusty. Can someone check my work and explain the missing parts?

• Domain

All Real

• Critical Points:

Find derivative:

$y' = -3x^2 + 6x + 9$

Set equal to 0:

$3x^2 - 6x - 9 = 0$

$(3x + 3)(x - 3) = 0$

$x = -1, x = 3$

Corresponding points: (-1, -3) & (3,30)

• Inflection Points:

Take second derivative:

$y'' = 6x - 6$

Set it to 0:

edit (fix typo)

$x = 1$

Corresponding point: (1, 13)

• Asymptotes

None

• Minor typo, for where second derivative is $0$ should have $x=1$. And this is not necessarily an inflection point, concavity must change. But it does. – André Nicolas Dec 12 '12 at 21:26
• @AndréNicolas Good catch, you have a good eye. – StrugglingWithMath Dec 12 '12 at 21:26

Yes, you find inflection points by taking the second derivative $y''$ and setting $y''$ equal to zero. Solve for x, to determine the point $(x, y)$ at which an inflection point may occur. (This procedure may not result in an inflection point, but in this case it does. If an inflection point exists, it will be at the point at which $y'' = 0$, but the converse is not always true.
For your critical points, find the corresponding $y$ values associated with your solutions fr $x$, respectively, to obtain the critical points: you want to write the points as ordered pairs.
• Note that your inflection point is when $6x = 6$ ---> $x = 1$ – Namaste Dec 12 '12 at 21:32