# Homework Help: P.O.I Question

1. Mar 29, 2007

### mathmann

1. The problem statement, all variables and given/known data
The slope of the curve y = ax(cubed) + bx(squared) + c is -12 at its point of inflection (2,-11), find a,b,c.

2. Relevant equations
f ' (x) = 3ax(squared) + 2bx
f " (x) = 3ax + 2b

3. The attempt at a solution

f ' (x) = 3ax(squared) + 2bx
-12 = 3a2(squared) + 2b(2)
-12 = 12a + 4b
b = -3a - 3

-12 = 3a(2)(sqrd) + 2(-3a -3)
a = -1

b = 0

-11 = ax(cubed) + bx(sqrd) + c
c = -3

2. Mar 29, 2007

### neutrino

That is correct.

Another way to do it, and probably the one "expected" from you, is to use the fact that at the point of inflection, f''(x)=0.

3. Mar 29, 2007

### mathmann

not exactly sure how to do that..

6ax + 2b = 0?

4. Mar 29, 2007

### neutrino

Yes, that's right. Therefore b = -6a. Solve for a.

5. Mar 29, 2007

### mathmann

so b = 6 not 0?

6. Mar 29, 2007

### neutrino

No, no!!
b = -6a is one relation, and is b = -3a - 3 is another. Now solve for a. Sorry for the confusion.

7. Mar 29, 2007

### Sourabh N

ur solution in OP is wrong . Check out step 5.

8. Mar 29, 2007

### mathmann

which do you consider step 5?

9. Mar 29, 2007

### Sourabh N

-12 = 3a(2)(sqrd) + 2(-3a -3)(?)
a 2 is missing in (?)

10. Mar 29, 2007

### mathmann

ok so a = 0 now

11. Mar 29, 2007

### neutrino

Sorry, my bad. As Sourabh has pointed out, your solution in the first post is in fact incorrect. a is not 0. What you would get from "step 5" is 0=0. That is because you got the relation bet. a and b from that very equation.

As I said earlier, use b = -6a and b = -3a -3 to solve for a.

12. Mar 29, 2007

### mathmann

-12=3ax(sqrd)+2(-6a)x
a=-1
b=6
c=-27

Last edited: Mar 29, 2007
13. Mar 29, 2007

### neutrino

Check again.

Once you get values for a, b and c, you can check if they are right by putting x=2 in the original equation and looking at the value you end up with.

14. Mar 29, 2007

### mathmann

ok im confused now.. ive found new values fro a(1),b(6) and c(5) but x doesnt = 2 when i enter it into the original equation.

Last edited: Mar 29, 2007
15. Mar 29, 2007

### Sourabh N

i think there's again a calculation mistake in #12.
-12 = 3a2(sqrd) + 2(-6a)2
-12 = -12a
a=1.

16. Mar 29, 2007

### mathmann

ok i got that one are the other two right? b=6 and c=5?

17. Mar 29, 2007

### neutrino

b = -6 (b = -6a, remember?)

18. Mar 29, 2007

### Sourabh N

no. b=-6a=-6

-11 = 1*2(cubed) - 6*2(sqrd) +c

19. Mar 29, 2007

### mathmann

a=1, b=-6, c =5

20. Mar 29, 2007

### mathmann

is this correct?

21. Mar 29, 2007

### pki15

You should get 3 equations:

$$12a + 2b = 0$$ (from 2nd derivative)
$$12a + 4b = -12$$ (from 1st derivative)
$$8a + 4b + c = -11$$ (from original equation)

Just solve for a,b,c from that info... I think you have all the right ideas, just getting wrong values.