The tangent to the curve y=e^2 x at the point (0, 1) meets x-axis at

We have,

$y=e^{2 x} \Rightarrow \frac{d y}{d x}=2 e^{2 x} \Rightarrow\left(\frac{d y}{d x}\right)_{(0,1)}=2$

The equation of the tangent at (0, 1) is

$y-1=2(x-0) \Rightarrow 2 x-y+1=0$

This meets x-axis at $(-1 / 2,0)$