An equation of the line tangent to the graph of f(x)=2x^3-3x^2 at the point of inflection is...?

Question

An equation of the line tangent to the graph of  f(x)=2x^3-3x^2 at the point of inflection is...?

pokemon23 · Answer

eashmore can you back to my history question I posted something after your done answering this person question :D

anonymous · Answer

First, let's find the first and second derivatives. We'll need both. $$f'(x) = 6x^2 - 6x$$$$f''(x) = 12x - 6$$Now, we need to find the inflection point. We find this by setting the second derivative to zero, and solving for x. 
We can use the x value from the inflection point and substitute this into the first derivative. This will be what we are looking for.

agentjamesbond007 · Answer

I got x= 1/2 to be the inflection point... from there, how do you find the equation of the line?

asnaseer · Answer

as @eashmore said above, if you substitute this value of x into the first derivative, you will get the slope of the line you are looking for.
then substitute the same x value into the original equation to get the 'y' value at that point.
you should then have the slope of the tangent line and a point (x,y) that it passes through. this should be enough for you to work out the equation of the tangent line at the point of inflection.