find a,b,c so y=ax^3+bx+c has turning pt at (2,-14), m=-9 at

[gtal3x]

Hello every one, i have this question to answer:
The curve y = ax^3+bx+c has turning points at (2,-14) and gradient of -9 when x = 1. Find the values of a, b and c [Ans: a=1,b=-12,c=2]

I have found a and b, but can not find c. Cant figure out how to do this... As I remember (might be wrong), to find c, I have to differentiate first, the problem is that c disappears ones differentiated. Thanks in advance for any help

 

[stapel]

The curve y = ax^3+bx+c has turning points at (2,-14)...

This is just one turning point: (x, y) = (2, -14). What is the other one?

I have found a and b, but can not find c.

Please reply showing your work. Thank you!

 

[gtal3x]

This is just one turning point: (x, y) = (2, -14). What is the other one?

Dont know, I coppied the question from the book...

Here is the work out for a and b:
y = ax^3+bx+c
y' = 3ax^2 + b
3a(2)^2+b=0
12a+b=0

3a(1)^2+b=-9
3a+b=-9 <--- multiply by -1 we get -3a-b=9

12a+b=0
-3a-b=9
------------
9a=9 -> a=1

12+b=0 -> b=-12

 

[stapel]

The curve y = ax^3+bx+c has turning points at (2,-14) and gradient of -9 when x = 1. Find the values of a, b and c

Your work so far looks good. The derivative is:

. . . . .dy/dx = 3ax[sup:3ma0gxpb]2[/sup:3ma0gxpb] + b

Since dy/dx = -9 at x = 1, then:

. . . . .-9 = 3a + b

The max/min point is obviously a critical point, so dy/dx = 0 at x = 2:

. . . . .0 = 12a + b

Solving the system of equations gives -9 - 3a = -12a, 9a = 9, and thus a = 1, so b = -12.

You now have one piece of information remaining: the fact that (x, y) = (2, -14) lies on the curve. You've got the original equation this far:

. . . . .y = x^3 - 12x + c

Now plug in the given point:

. . . . .-14 = 8 - 24 + c

Solve for c. :wink:

 

[gtal3x]

thanks a lot, i got it now