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If f(x) is a polynomial function of the second degree such that f(-3) = 6, f(0) = 6 and f(2) = 11, then the graph of the function f(x) cuts the ordinate at x = 1 at the point

A

(1, 8)

B

(1, 4)

C

(-2, 1)

D

(1, 9)

Text Solution

Verified by Experts

The correct Answer is:
A
Let `f(x)=ax^(2) +bx +c`
`f(0) =6 rArr c=6`
`f(-3)=6 rArr 9a-3b+6=6rArr b=3a`
`f(2)=11 rArr 4a +2b +6=11 rArr 10a =5 rArr a=(1)/(2)`
`rArr b =(3)/(2) and c =6`
`f(x)=(1)/(2) x^(2)+(3)/(2)x +6`
At `x =1`
`f(1)=(1)/(2) xx 1 +(3)/(2) xx1 +6=8`
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