If the curve y=ax2+bx+c passes through the point (1, 2) such that the slope of the tangent at the origin is equal to 1, then (a, b, c) =

It is given that $y = a x^{2} + b x + c$ passes through $(1, 2)$
and $(0, 0)$ . Therefore,

$2 = a + b + c$ and $0 = c \Rightarrow a + b = 2$ and $c = 0$

Now, $\begin{aligned} y = a x^{2} + b x + c \\ \Rightarrow \frac{d y}{d x} = 2 a x + b \Rightarrow {(\frac{d y}{d x})}_{(0, 0)} = b \end{aligned}$

But, ${(\frac{d y}{d x})}_{(0, 0)} = 1$

$\Rightarrow b = 1$

But, $a + b = 2 .$ Therefore, $a = 1$ . Hence, $(a, b, c) = (1, 1, 0) .$

If the curve $y = a x^{2} + b x + c$ passes through the point (1, 2) such that the slope of the tangent at the origin is equal to 1, then $(a, b, c) =$