# Thread: Stationary Points Possible Values

1. ## Stationary Points Possible Values

Hi,

I am having trouble with part d) of this question. It follows on with other parts of a question which I have attached. I have written that 'p' can indeed have stationary points but am not sure what the possible values of 'p' could be. If anyone can list these possible values that would be great, as this is apart of a big assignment that is due soon. Please note this is from question 4 from the first picture.

Thank You and Kind Regards.

2. ## Re: Stationary Points Possible Values

To be clear, part (d) asks about the number, p, of possible stationary points of the quartic polynomial function $y=ax^4+bx^3+cx^2 +dx+e$

Stationary points occur where $\dfrac{dy}{dx} = 0$. The derivative of a quartic polynomial function is a cubic polynomial function.

For such a cubic function, what are the possible number of zeros?

3. ## Re: Stationary Points Possible Values

Hi,

Would the answer just be P = 1, 2 or 3 given the points we have been given of where it passes

4. ## Re: Stationary Points Possible Values

Originally Posted by BAH0003
Hi,

Would the answer just be P = 1, 2 or 3 given the points we have been given of where it passes
Note that part (d) has no relation to parts (a), (b), and (c) and the given points.

The polynomial in parts (a), (b), and (c) is $y=ax^2+bx^3+cx^2+d$. There is no linear term.

The polynomial in part (d) is $y=ax^2+bx^3+cx^2+dx + e$. Not the same as above.

5. ## Re: Stationary Points Possible Values

Is it okay if you could tell me the answer, I'm in kind of a short space of time to get this done, and it is the only thing I have left.

6. ## Re: Stationary Points Possible Values

Originally Posted by BAH0003
Is it okay if you could tell me the answer
No. I'm sorry, but that's Forum policy.

In post #2 Cervesa asked you how many points you would have for a cubic. Do you know how to find the answer to that question? (Hint: You want to set dy/dx = 0. What kind of polynomial equation do you get for this condition?)

-Dan