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..

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

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

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!