"y=(1+2x)^10" Is this the equation of the curve??
First find what is dy/dx.
Note that the gradient of the tangent to the curve at x=0 is equal to the gradient of the curve at the pt. x=0
To find dy/dx, use chain rule.
dy/dx=10{(1+2x)^9} x 2
=20{(1+2x)^9}
Then sub x=0 into dy/dx.
U will get dy/dx=20
Eqn.of tangent to the curve at x=0
=> [(y-1)/x] = 20
Complete the rest on your own...