# Can not find critical points for turning points.

• Oct 5th 2012, 03:20 PM
astuart
Can not find critical points for turning points.
Is there any possible way that the critical points for an equation can be found out short of find when the equation = 0, or is not defined?

The equation I have is

$\displaystyle 3x^2 - 2x - 6$

I can't figure out how to factor it, and it's also just undefinable if it's put through the quadratic equation.

So I have no idea how I'm supposed to find the critical points. I know there has to be two turning points seeing as it's third degree, but can't figure out what I should be doing..
• Oct 5th 2012, 03:40 PM
TheEmptySet
Re: Can not find critical points for turning points.
Quote:

Originally Posted by astuart
Is there any possible way that the critical points for an equation can be found out short of find when the equation = 0, or is not defined?

The equation I have is

$\displaystyle 3x^2 - 2x - 6$

I can't figure out how to factor it, and it's also just undefinable if it's put through the quadratic equation.

So I have no idea how I'm supposed to find the critical points. I know there has to be two turning points seeing as it's third degree, but can't figure out what I should be doing..

I don't know what you mean undefineable in the quadratic formula it gives

$\displaystyle x=\frac{2 \pm \sqrt{4-4(3)(-6)}}{6}=\frac{1 \pm \sqrt{19}}{3}$
• Oct 5th 2012, 03:55 PM
astuart
Re: Can not find critical points for turning points.
Quote:

Originally Posted by TheEmptySet
I don't know what you mean undefineable in the quadratic formula it gives

$\displaystyle x=\frac{2 \pm \sqrt{4-4(3)(-6)}}{6}=\frac{1 \pm \sqrt{19}}{3}$

Well it certainly does. I forgot the minus sign, so was getting a negative underneath the square root.

Thanks!