If the curve y = ax^{2} + bx + c, x in R, pas | vedclass

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(C) Given the curve $y = ax^{2} + bx + c$. Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation: $0

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$a = 1, b = 1, c = 0$

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(C) Given the curve $y = ax^{2} + bx + c$. Since the curve passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation: $0 = a(0)^{2} + b(0) + c$,which gives $c = 0$. The derivative of the curve is $\frac{dy}{dx} = 2ax + b$. The tangent line at the origin $(0, 0)$ is $y = x$,which has a slope of $1$. Thus,$\left. \frac{dy}{dx} \right|_{(0,0)} = 2a(0) + b = 1$,which gives $b = 1$. Since the curve passes through the point $(1, 2)$,we substitute $x = 1, y = 2, b = 1, c = 0$ into the equation: $2 = a(1)^{2} + 1(1) + 0$. This simplifies to $2 = a + 1$,so $a = 1$. Therefore,the values are $a = 1, b = 1, c = 0$.

If the curve $y = ax^{2} + bx + c, x \in R,$ passes through the point $(1, 2)$ and the tangent line to this curve at the origin is $y = x$,then the possible values of $a, b, c$ are:

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