If the line y = 2x touches the curve $y=ax^2+bx+c$ at the point where x = 1 and the curve passes through the point (-1, 0), then

$a=\frac{1}{2}, b=1, c=\frac{1}{2}$

We have,

$y=a x^2+b x+c$               ........(i)

$\Rightarrow \frac{d y}{d x}=2 a x+b \Rightarrow\left(\frac{d y}{d x}\right)_{x=1}=2 a+b$

Since, the line y = 2x touches (i) at the point where x = 1.

∴ (Slope of the tangent at x = 1) = (Slope of the line y = 2x)

$\Rightarrow 2 a+b=2$               ........(ii)

Putting x = 1 in y = 2x, we get y = 2

Thus the given line touches the curve (i) at (1, 2)

Clearly, (1, 2) and (-1, 0) lie on (i). Therefore,

$2=a+b+c$        ...(iii)          and,   $0=a-b+c$         .......(iv)

Solving (i), (ii) and (iii), we get $a=\frac{1}{2}, b=1$ and $c=\frac{1}{2}$