#1
Yep, need maths help again lol

this time its about differentiations:

the graph of y=ax^3 + bx has gradient 4 when x=-1 and gradient 31 when x=2.
find the values of a and b.

Ive done the whole exercise except this question and i need help lol.
#2
Never did Gradient, is that related to slope or derivative?
"Virtually no one who is taught Relativity continues to read the Bible."

#4
differentiate y

dy/dx = ax^2+b
m = ax^2 +b
when x = -1, m =4
when x = 2, m = 31

simultaneous equation

4 = a + b
31 = 4a + b

Solve 1st and 2nd equation

3a = 27
a = 9

b = 4-9
b = -5


I hope that's right
#5
Quote by KIDRoach
differentiate y

dy/dx = ax^2+b
m = ax^2 +b
when x = -1, m =4
when x = 2, m = 31

simultaneous equation

4 = a + b
31 = 4a + b

Solve 1st and 2nd equation

3a = 27
a = 9

b = 4-9
b = -5


I hope that's right


thanks mate, except when you differentiate y it becomes 3ax^2 + b

But thank you very much! I didnt think of simultaneous equations lol
#6
m = dy/dx = 3ax^2 + b

4 = 3a + b

b = 4 - 3a

31 = 6a + b

31 = 6a + 4 - 3a

31 = 3a + 4

3a = 27

a = 9

b = 4 - 3a

b = 4 - 27

b = -23


Different to the guy above, but I'm quite sure dy/dx = 3ax^2 and not just ax^2.


EDIT: Yeah, I'm definitely right.
Co-President of UG's Tubgirl Virgins Club

Last edited by JamieB at Nov 14, 2007,