1. ## Help please how to do this?

f(x) = ax˛ + bx + c. If f(1) = 6, f (-1) = 8, f(-2) = 18, find f (2)
Can someone please do the solution for me. I know that the answer is 14 but I dont know how to get that.
Thanks!

2. Originally Posted by Detanon
f(x) = ax˛ + bx + c. If f(1) = 6, f (-1) = 8, f(-2) = 18, find f (2)
Can someone please do the solution for me. I know that the answer is 14 but I dont know how to get that.
Thanks!

$\displaystyle f(1) = a + b + c =6$
$\displaystyle f(-1) = a -b + c = 8$
$\displaystyle f(-2) = 4a-2b+c = 18$

You have three equations and three variables

3. Wait so if i put
f(2) = 4a + 2b + c
then what do I do?
Is that the right method?
Thanks

4. Originally Posted by Detanon
Wait so if i put
f(2) = 4a + 2b + c
then what do I do?
Is that the right method?
Thanks
You're meant to get the values of a, b and c first by solving the equations given to you in post #2.

5. Originally Posted by Detanon
f(x) = ax˛ + bx + c. If f(1) = 6, f (-1) = 8, f(-2) = 18, find f (2)
Can someone please do the solution for me. I know that the answer is 14 but I dont know how to get that.
Thanks!
the first time, you need find a, b, c:
we have $\displaystyle f(1)$so $\displaystyle x=1$ and change into the $\displaystyle f(x)=ax^2+bx+c$
then $\displaystyle f(-1) and f(-2)$ is same the$\displaystyle f(1)$

so we have

$\displaystyle a+b+c=6$
$\displaystyle a-b+c=8$
$\displaystyle 4a-2b+c=18$
you can find a, b,c
then change a, b, c you found go into $\displaystyle f(x)=ax^2+bx+c$
finally $\displaystyle f(2)$ you can find easily.
if you didnot understand you can ask me and i can explain again. sorry my english bad something i wrote wrong .

6. You can find b by: $\displaystyle f(1)-f(-1)$

7. Originally Posted by e^(i*pi)
You can find b by: $\displaystyle f(1)-f(-1)$
correct i did not remember this. thanks khakha

8. Thanks, yo've helped me alot.

Originally Posted by e^(i*pi)
You can find b by: $\displaystyle f(1)-f(-1)$
wait, so does that mean
6 - 8 then.
so b is -2?

9. Originally Posted by Detanon
wait, so does that mean
6 - 8 then.
so b is -2?
No, look again and remember that if we take away a negative we must add it.

ie b-(-b) = 2b

10. Originally Posted by Detanon
wait, so does that mean
6 - 8 then.
so b is -2?
$\displaystyle f(1)-f(-1)=(a+b+c)-(a-b+c)=6-8$
$\displaystyle a+b+c-a+b-c=-2$
$\displaystyle 2b=-2$
$\displaystyle b=-1$