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4th degree polynomial question

15082743735961080891215.jpgI have literally no idea where to begin the question
Reply 1
Just sketch it, the first derivative is 12x^3-12x^2-24x =12x(x^2-x-2) which is 0 when x= 0 or x^2 - x - 2 = (x+1)(x-2) = 0 so x=0, x = -1 and x=2 are the stationary points. What are the values?

That should give you enough to sketch it, since you know that it has to go to +infinity as |x| \to +\infty. Then just see what lines y=k gives rise to 4 intersections.
Reply 2
Original post by Zacken
Just sketch it, the first derivative is 12x^3-12x^2-24x =12x(x^2-x-2) which is 0 when x= 0 or x^2 - x - 2 = (x+1)(x-2) = 0 so x=0, x = -1 and x=2 are the stationary points. What are the values?

That should give you enough to sketch it, since you know that it has to go to +infinity as |x| \to +\infty. Then just see what lines y=k gives rise to 4 intersections.

Thank you :smile:

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