If f(x) = x^2 + bx + c and f(1+k) = f(1-k) fo | vedclass

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(A) Given $f(x) = x^2 + bx + c$. Since $f(1+k) = f(1-k)$ for all $k \in R$,the axis of symmetry of the parabola is $x = 1$. The formula for the axis o

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$f(1) < f(0) < f(-1)$

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(A) Given $f(x) = x^2 + bx + c$. Since $f(1+k) = f(1-k)$ for all $k \in R$,the axis of symmetry of the parabola is $x = 1$. The formula for the axis of symmetry of $f(x) = ax^2 + bx + c$ is $x = -b/(2a)$. Here $a = 1$,so $-b/2 = 1$,which implies $b = -2$. Thus,$f(x) = x^2 - 2x + c$. Now,calculate the values: $f(0) = 0^2 - 2(0) + c = c$. $f(1) = 1^2 - 2(1) + c = 1 - 2 + c = c - 1$. $f(-1) = (-1)^2 - 2(-1) + c = 1 + 2 + c = c + 3$. Comparing these values: $c - 1 Therefore,$f(1) < f(0) < f(-1)$.

If $f(x) = x^2 + bx + c$ and $f(1+k) = f(1-k)$ for all $k \in R$,for two real numbers $b$ and $c$,then:

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