Find a polynomial f(x) of degree 2 where f(0) | vedclass

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(D) Let the polynomial be $f(x) = ax^{2}+bx+c$.Given $f(0)=8$,we have $a(0)^{2}+b(0)+c=8$,which implies $c=8$.Thus,the polynomial is $f(x) = ax^{2}+bx

Vedclass · Accepted Answer

$x^{2}+3x+8$

Vedclass · Answer

(D) Let the polynomial be $f(x) = ax^{2}+bx+c$.Given $f(0)=8$,we have $a(0)^{2}+b(0)+c=8$,which implies $c=8$.Thus,the polynomial is $f(x) = ax^{2}+bx+8$.Given $f(1)=12$,we have $a(1)^{2}+b(1)+8=12$,which simplifies to $a+b=4$ (Equation $i$).Given $f(2)=18$,we have $a(2)^{2}+b(2)+8=18$,which simplifies to $4a+2b=10$,or $2a+b=5$ (Equation $ii$).Subtracting Equation $(i)$ from Equation $(ii)$,we get $(2a+b)-(a+b) = 5-4$,which gives $a=1$.Substituting $a=1$ into Equation $(i)$,we get $1+b=4$,which gives $b=3$.Therefore,the required polynomial is $f(x) = x^{2}+3x+8$.

Find a polynomial $f(x)$ of degree $2$ where $f(0)=8, f(1)=12, f(2)=18$.

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