Find an eqn of the stright line that is tangent to the graph of f(x) = sqr(x+1) and parallel to x-6y+4=0?

Question

This is the only homework question i slumped aganist and well it obviously has to do with the derivative function because that's what we're learning...if think you are able to answer it please do or just show me how to do it or what to do please?

Answer 1

x-6y+4=0 solve for y=(-4 -x)/6 = -(1/6)x - (2/3)
slope = -(1/6) parallel lines have the same slope

now find where on the line f(x)= sqr(x+1) = -(1/6)
find f'(x) which I don't recall how to do but it sounds like you should if you did all the other problems and set f'(x) = -(1/6)