here's the problem

Find the equations of the tangent line & normal line to the graph of the equation at the indicated point..

____

y = √x+1; (3,2)

i got the slope and its

__1___

2√x+1

so i have to plug it in the tangent line & normal line formula which is

$\displaystyle

y - y1 = m(x-x1) - tangent line$

and

$\displaystyle

y - y1 = 1/m(x-x1) - normal line$

so its goin' to be:

y-2 = 1/2√x+1 (x-3)

and my answer is:

___ ___

0 =1/2√x+1 + 1/6√x+1

pls help..thanx