Graph the function 3x^3-2x^2-2x+1

Question

anonymous · Answer

just use a graphing calculator

imstuck · Answer

It's a cubic so it will look weird.  Something like this, actually..|dw:1405230310550:dw|

anonymous · Answer

Do you know derivatives?

anonymous · Answer

Not really @doulikepiecauseidont

anonymous · Answer

So, would that be a yes or you have no idea what I'm talking about? Lol it would help if you cleared that up

imstuck · Answer

The far left x intercept is at -.768, the second one is at .434, and the third one is at 1

anonymous · Answer

How did you get the points @IMStuck

imstuck · Answer

I used my graphing calculator!  It's a wonderful thing!  The only thing I knew for sure was that it was a cubic because it was to the 3rd power.  Use the calculator!

anonymous · Answer

Okay!

imstuck · Answer

@doulikepiecauseidont  how will the derivative help?  Please clarify...for my benefit.

imstuck · Answer

9x^2-4x-2 is the first derivative, but you don't know any points on the line for sure to find the slopes...

anonymous · Answer

Ok, well if you were to take the first and second derivative, it would help describe the behavior of the graph

anonymous · Answer

Set the first derivative=0 and get those points called critical points which are the points that have a slope of 0

imstuck · Answer

9x^2-4x-2=0.  Then factor it to find the critical points...

imstuck · Answer

How do you find the critical points?

anonymous · Answer

Plug the critical points into the original function and plot those. Then you split the graph into regions bounded by the critical points and you plug in points from the left and the right of each point and get the signs of each, + corresponds to an increasing graph and - to a decreasing graph. 

Once you have those, get the second derivative and set it=0 to get the inflection points (where the graph changes from concave down to concave up) and then with that accordingly you can get a good estimate too what the graph behaves like

anonymous · Answer

@IMStuck The critical points are where f'(x) (derivative) are equal to 0 or it doesn't not exist (in this case of apolynomial all x-values will work but in a rational function 2x/(3x-6) then you'd have one there