The curve y = ax^2 + bx + c passes through th | vedclass

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(A) Given the curve $y = ax^2 + bx + c$. Since it passes through $(1, 2)$,we have $2 = a(1)^2 + b(1) + c$,so $a + b + c = 2$ ... $(1)$. Since it passe

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$\frac{1}{24}$

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(A) Given the curve $y = ax^2 + bx + c$. Since it passes through $(1, 2)$,we have $2 = a(1)^2 + b(1) + c$,so $a + b + c = 2$ ... $(1)$. Since it passes through the origin $(0, 0)$,we have $0 = a(0)^2 + b(0) + c$,so $c = 0$. Substituting $c = 0$ into $(1)$,we get $a + b = 2$ ... $(2)$. The tangent at the origin is $y = x$,which means the slope of the tangent at $x = 0$ is $1$. Since $\frac{dy}{dx} = 2ax + b$,at $x = 0$,$\frac{dy}{dx} = b$. Thus,$b = 1$. From $(2)$,$a + 1 = 2$,so $a = 1$. The curve is $y = x^2 + x$. The minima occurs where $\frac{dy}{dx} = 2x + 1 = 0$,which gives $x = -\frac{1}{2}$. The area bounded by the curve $y = x^2 + x$,the tangent $y = x$,and the ordinate $x = -\frac{1}{2}$ is given by: $A = \int_{-\frac{1}{2}}^{0} (x - (x^2 + x)) \, dx = \int_{-\frac{1}{2}}^{0} (-x^2) \, dx$. $A = \left[ -\frac{x^3}{3} ight]_{-\frac{1}{2}}^{0} = 0 - \left( -\frac{(-\frac{1}{2})^3}{3} ight) = -\frac{1}{8 	imes 3} = -\frac{1}{24}$. Since area is positive,$A = \frac{1}{24} 	ext{ sq. units}$.

The curve $y = ax^2 + bx + c$ passes through the point $(1, 2)$ and its tangent at the origin is the line $y = x$. The area bounded by the curve,the ordinate of the curve at its minima,and the tangent line is

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