# Math Help - Finding a horizontal tangent

1. ## Finding a horizontal tangent

I have a homework question I cannot solve. The question is:

For what values does the graph of f(x) x^3 + 3x^2 + x + 3 have a horizontal tangent.

I though I could take the derivative of this function and set that to zero and solve using the quadratic formula but that does not seem to work. Does anyone have any hints to lead me to the right solution?

EDIT: Can someone move this thread to the urgent homework help? I posted this in the wrong section

2. Originally Posted by fishguts
I have a homework question I cannot solve. The question is:

For what values does the graph of f(x) x^3 + 3x^2 + x + 3 have a horizontal tangent.

I though I could take the derivative of this function and set that to zero and solve using the quadratic formula but that does not seem to work. Does anyone have any hints to lead me to the right solution?

EDIT: Can someone move this thread to the urgent homework help? I posted this in the wrong section
Your description how you would solve the question is correct. I assume that you have made a minor mistake.

$f(x)=x^3+3x^2+x+3~\implies~f'(x)=3x^2+6x+1$

Use the quadratic formula to solve f'(x) = 0:

$x=\dfrac{-6\pm\sqrt{36-4\cdot 3\cdot 1}}{2\cdot 3}=-1\pm\frac13\sqrt{6}$

That means $x \approx -0.183...~\vee~x\approx -1.816...$

3. Originally Posted by earboth
Your description how you would solve the question is correct. I assume that you have made a minor mistake.

$f(x)=x^3+3x^2+x+3~\implies~f'(x)=3x^2+6x+1$

Use the quadratic formula to solve f'(x) = 0:

$x=\dfrac{-6\pm\sqrt{36-4\cdot 3\cdot 1}}{2\cdot 3}=-1\pm\frac13\sqrt{6}$

That means $x \approx -0.183...~\vee~x\approx -1.816...$
Thanks! I see where I made my mistake at