High school calculus tangent equation problem

determine an equation of the line that is tangent to the graph f(x) = (x+1)^1/2 and parallel to x -6y+4 = 0

First, find the derivative of x - 6y + 4 = 0

x + 4 = 6y
(1/6) * x + 2/3 = y

So we need to know when f'(x) = 1/6

f(x) = (x + 1)^(1/2)
f'(x) = (1/2) * (x + 1)^(-1/2)
1/6 = (1/2) * (x + 1)^(-1/2)
1/3 = (x + 1)^(-1/2)
3 = (x + 1)^(1/2)
9 = x + 1
8 = x

Now, find what f(8) is:

f(8) = (8 + 1)^(1/2) = 9^(1/2) = 3

So we need a line with a slope of 1/6 that passes through (8 , 3)

y - 3 = (1/6) * (x - 8)
y - 3 = (1/6) * x - 4/3
y = (1/6) * x - 4/3 + 9/3
y = (1/6) * x + (5/3)

There you go.