Functions! f(x)=

• Jan 5th 2009, 02:21 AM
coyoteflare
Functions! f(x)=
ok so i have my core 1 exam on friday and am struggling with SOME functions.

i am at work and dont have the question for reference but its alogn the lines of

(x-2) is a factor of f(0)

find the values of k and c in f(x)=2x^2+kc+c

• Jan 5th 2009, 02:48 AM
craig
I'm not sure but did you mistype your question, I think it should be something like:
$f(x)=2x^2+kx+c$

Can you remember what other information you are given about the roots of the equation?

With questions like these you usually use the discriminant, $b^2-4ac$.

For equal roots, $b^2-4ac=0$
For diffferent real roots, $b^2-4ac >0$
For no real roots, $b^2-4ac<0$

I know this doesn't answer your question but hope it helps you when you have the actual question in front of you (Wink)
• Jan 5th 2009, 02:54 AM
coyoteflare
information
it very possibly said somethign along the lines of f(x)=6

if that doesn't mean anything i will have al ook when i get home and retype on this post with the whole question and how it is worded
• Jan 5th 2009, 02:59 AM
craig
Yeh post the actual question when you get home and I'll try and get my head round it then ;)
• Jan 5th 2009, 09:11 AM
coyoteflare
the actual question!
You are given that f(x) = x^3 + kx +c. The value of f(0) is 6, and x-2 is a factor of f(x).

find the values of k and c

there we go, helps to have what your asking in front of you lol, thanks in advance :D
• Jan 5th 2009, 12:43 PM
mr fantastic
Quote:

Originally Posted by coyoteflare
You are given that f(x) = x^3 + kx +c. The value of f(0) is 6, and x-2 is a factor of f(x).

find the values of k and c

there we go, helps to have what your asking in front of you lol, thanks in advance :D

f(0) = 6 => 6 = c.

(x - 2) is a factor => f(2) = 0 => 0 = 2^3 + 2k + c

And now you should be able to finish this.
• Jan 5th 2009, 12:47 PM
coyoteflare
thanks, can you explain to me why c is 6? is it to do with >0 and <0 ?

0 = 2^3 + 2k + 6

0 = 8 + 2k + 6

-14 = 2k

-7 = k ?

thanks again
• Jan 5th 2009, 12:51 PM
mr fantastic
Quote:

Originally Posted by coyoteflare
thanks, can you explain to me why c is 6? is it to do with >0 and <0 ?

0 = 2^3 + 2k + 6

0 = 8 + 2k + 6

-14 = 2k

-7 = k ? Mr F says: yes.

thanks again

Quote:

f(x) = x^3 + kx +c. The value of f(0) is 6
Therefore 6 = 0^3 + k(0) + c ......
• Jan 5th 2009, 12:51 PM
craig
Ah this isn't as hard as I first thought :)

You have the following equation

$x^3 + kx +c$

and you are told that $f(0) = 6$

You can therefore work out that then you sub 0 into the equation it equals 6, giving you:

$0^3 + 0k +c = 6$

therefore $c = 6$

Reading on from this you are told that x - 2 is a factor, therefore subbing 2 into the equation will give you zero.

$2^3 + 2k + 6 = 0$

The rest I'm sure is clear :)

Hope this explains it

Craig
• Jan 5th 2009, 12:52 PM
craig
Sorry I've got beaten to the answer ;)
• Jan 5th 2009, 12:57 PM
coyoteflare
Thanks guys!
not as hard as i thought, just never had functions explained to me an got confused xD

thanks again!