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i cant find a way to isolate a or b to solve

have you found the derivative?

the derivative is 3ax^2 + 2bx but that doesnt help solve it

f(x) = ax^3 + bx^2
f(2) = 8a + 4b = -4
f(3) = 27a + 9b = 99

Can you now solve for a, b ? @Lachlan1996

wait wait, hold on youve lost me here.

lol, take your time. Sorry :)

uhhh i got 27a + 6b =99 but that is the gradient as in the derivative

so i dont think you can simultaneously solve it

because you are simultaneously solving a derivative and a point

but he is solving a f' (x) with a f (x)

a and b will be the same in the original and the derivative

ah yes i see

in the above equation isnt it meant to be 6a not 9a though?

wait no i mean 6b not 9b

27a+6b=99
8a+ 4b=-4

uhh okie doke so a=0.5b-0.5

however substituting that into 27a + 6b = 99 i got 8.333333 as being b

I got a=7, b=-15

oh no made an error forgot to subtract 6b

OH MY FAULT , sorry :(

this seems like alot of work just to start off the question, do you think this would be in an exam?

Depends but I do think, Yep! :)

For calculus, it doesn't seem out of bounds for a test.

ahh god, in a year 11 maths exam? i think im buggered then

Whether it will come or not in exam but I think you must be PREPARED

mmm yes my exam is in 5 weeks.

Creating a small system and solving it with substitution is fairly standard in calculus.

You have much time for it, go through these type of questions and ask your problems here...

okie doke, thanks very much for the help mate, its much appreciated. thankyou to you both

You're welcome.

Welcome! :)

uhhhh small problem

we got the answer wrong? its meant to be y=7x^3 -15x^2

Try to do it again, I think calculation mistake?

yes... a=7, b= -15

ah yes, last step of my working didnt put the -ve in the 15

hmn, Take care of these small mistakes .

whew! you scared me ;)

will do, doesnt help when you get frustrated but. maths is not easy for me

I think it is f'(3)=99 not f(3)=99