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 April 8th, 2013, 10:12 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Maximum and Minimum Turning Points I've found 'c' but I'm having trouble with finding the last 2 variables. Can anyone help me out here? Many thanks. Q. The function $y=ax^3+bx^2+c$ has turning points at (0, 4) & (-1, 5). Find the values of a, b & c. Attempt: $4=a(0)^3+b(0)^2+c\rightarrow c=4$ for (0, 4). $5=a(-a)^3+b(-1)^2+4\rightarrow-a+b=1\rightarrow b=1+a$ for (-1, 5). $\frac{dy}{dx}=3ax^2+2bx$ $\frac{d^2y}{dx^2}=6ax+2b=0\rightarrow3ax+b=0\right arrow3a(-1)+1+a=0\rightarrow a=\frac{1}{2}$ Ans.: (From text book): a = 2, b= 3, c = 4.
 April 8th, 2013, 10:50 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,842 Thanks: 1068 Math Focus: Elementary mathematics and beyond Re: Maximum and Minimum Turning Points $3ax^2\,+\,2bx\,=\,0\,\text{ if }x\,=\,-1 \\ 3a\,-\,2b\,=\,0 \\ a\,=\,\frac23b \\ ax^3\,+\,bx^2\,+\,4\,=\,5\text{ if }x\,=\,-1 \\ -\frac23b\,+\,b\,=\,1\,\Rightarrow\,b\,=\,3 \\ 3a\,-\,2(3)\,=\,0\,\Rightarrow\,a\,=\,2$
 April 8th, 2013, 11:15 AM #3 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Maximum and Minimum Turning Points Thank you.

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