Question

Suppose f(x) is a quadratic polynomial satisfying

f(0)=1
f(4)−f(2)=2
f(6)−f(4)=3
What is the value of f(8)?

Answer 1

let the quadratic polynomial be \(\large f(x)=ax^2+bx+c\)
from f(0)=1, you can easily find the value of 'c', which is ?

Answer 2

1?

Answer 3

correct!
so, f(x) = ax^2+bx+1

now can you find, f(2), f(4), f(6) ??

the goal is to use
f(4)−f(2)=2
f(6)−f(4)=3
and get 2 equations in a and b, so that they can be solved simultaneously, to get the values of a and b

Answer 4

ok i'm lost :'(

Answer 5

ok, can you find f(2) ??

Answer 6

um no

Answer 7

@ziko1995 let him do that work of getting equations

Answer 8

@hartnn ok :)

Answer 9

ok, to find f(2), just plug in x= 2 in f(x)
so, what u get after u put x=2 in ax^2+bx+1 ?

Answer 10

4a + 4b + 1

Answer 11

@killua_vongoladecimo It's 4a+2b+1 cause we haven't \[bx ^{2}\] :)

Answer 12

sorry

Answer 13

x=2 so, bx, gives 2x, right ??
so, its f(2)=4a+2b+1
similarly find f(4)

Answer 14

hmm.. in
f(4) - f(2) = 2

after using ax^2 + bx + 1

it will become
(16a + 4b + 1) - (4a + 2b + 1) = 2

then when simplified

12a + 2b = 2

right?

Answer 15

correct! simplify it.
now do f(6)-f(4)=3

Answer 16

20a + 2b = 3

Answer 17

right? what's next?

Answer 18

@hartnn

Answer 19

12a+2b=2
20a+2b=3

just subtract one equation from other and b gets eliminated!

Answer 20

so
a= 1/8
b= 1/4
?

Answer 21

correct!
so, we got our f(x)!
\(\huge f(x)=(1/8)x^2+(1/4)x+1\)

to get f(8), just plug in x=8 there :)

Answer 22

oh... so f(8) = 11?

Answer 23

yes, i get 11 too :)

Answer 24

oh so that's how it is.....thank you hartnn but i still got more questions haha...

Answer 25

welcome ^_^
sure! ask :)
better you close this first and ask in new post ?