A quadratic polynomial y = f(x) with absolute | vedclass

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(A) Let the quadratic polynomial be $f(x) = ax^2 + bx + c$. Given $c = 3$ and $a = 1$,so $f(x) = x^2 + bx + 3$.Since the curve is symmetric about $x =

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$4$

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(A) Let the quadratic polynomial be $f(x) = ax^2 + bx + c$. Given $c = 3$ and $a = 1$,so $f(x) = x^2 + bx + 3$.Since the curve is symmetric about $x = 1$,the vertex occurs at $x = 1$.The derivative $f'(x) = 2x + b$. Setting $f'(1) = 0$ gives $2(1) + b = 0$,so $b = -2$.Thus,the polynomial is $f(x) = x^2 - 2x + 3$.The area bounded by the curve $y = f(x)$ and the line $y = 3$ is found by setting $f(x) = 3$:$x^2 - 2x + 3 = 3$ $\Rightarrow x^2 - 2x = 0$ $\Rightarrow x(x - 2) = 0$.The intersection points are $x = 0$ and $x = 2$.The area is given by the integral $\int_{0}^{2} (3 - f(x)) dx = \int_{0}^{2} (3 - (x^2 - 2x + 3)) dx = \int_{0}^{2} (2x - x^2) dx$.Evaluating the integral: $[x^2 - \frac{x^3}{3}]_{0}^{2} = (4 - \frac{8}{3}) - 0 = \frac{12 - 8}{3} = \frac{4}{3}$.

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