If f(x)=ax²+bx+c and f(1)=6,f(2)=11,f(3)=18 find the value of a, b,c

f(x) = ax² + bx + c

f(1) = 6

f(2) = 11

f(3) = 18

value of a,b,c

Substitute x = 1 , 2 , 3 in the given polynomial.

★ f(1) = a(1)² + b(1) + c

⟹ 6 = a + b + c -- equation (1)

Similarly,

★ f(2) = a(2)² + b(2) + c

⟹ 11 = 4a + 2b + c -- equation (2)

★ f(3) = a(3)² + b(3) + c

⟹ 18 = 9a + 3b + c -- equation (3)

Subtract equation (1) from (2).

⟹ 4a + 2b + c - (a + b + c) = 11 - 6

⟹ 4a + 2b + c - a - b - c = 5

⟹ 3a + b = 5 -- equation (4).

From equation (3) ,

⟹ 18 - c = 9a + 3b

⟹ 18 - c = 3(3a + b)

Substitute the value of 3a + b from equation (4).

⟹ 18 - c = 3 * 5

⟹ 18 - 15 = c

⟹ 3 = c

substitute the value of c in equation (1).

⟹  6 = a + b + 3

⟹ 6 - 3 = a + b

⟹ 3 = a + b -- equation (5)

Subtract equation (5) from equation (4).

⟹ 3a + b - (a + b) = 5 - 3

⟹ 3a + b - a - b = 2

⟹ 2a = 2

⟹ a = 1

Substitute a value in equation (5).

⟹ 1 + b = 3

⟹ b = 3 - 1

⟹ b = 2

Therefore,

a = 1

b = 2

c = 3.

I hope it will help you.

Regards.

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^