Question 117783
Here is the graph of your original line, y=2x+1, and the point of interest. 
{{{drawing( 300, 300, -12, 12, -12, 12,grid( 1 ),circle(-4,3,.4),graph( 300, 300, -12,12, -12, 12, 2x+1))}}}
Parallel lines have the same slope.
Use the point slope form of a line.
{{{y-y[1]=m(x-x[1])}}}
{{{y-3=2(x-(-4))}}}
{{{y-3=2(x+4)}}}
{{{y-3=2x+8}}}
{{{y=2x+11}}}
{{{drawing( 300, 300, -12, 12, -12, 12,grid( 1 ),circle(-4,3,.4),graph( 300, 300, -12,12, -12, 12, 2x+1,2x+11))}}}