# Thread: Functions! f(x)=

1. ## 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

2. 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

3. ## 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

4. Yeh post the actual question when you get home and I'll try and get my head round it then

5. ## 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

6. 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
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.

7. 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

8. 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
f(x) = x^3 + kx +c. The value of f(0) is 6
Therefore 6 = 0^3 + k(0) + c ......

9. 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

10. Sorry I've got beaten to the answer

11. ## Thanks guys!

not as hard as i thought, just never had functions explained to me an got confused xD

thanks again!