Question: Find an equation of the straight line that is tangent to the graph of f(x) = √x+1 and parallel to x-6y+4=0. I've found f`(x). When I write lim, assume "h->0" (as h approaches 0) is beneath it. I'd like to use LaTex to show you, but it seems too confusing to figure out. f`(x)=lim f(x+h) - f(x) / h f`(x)=lim √x+h+1 - √x+1 / h x √x+h+1 + √x+1 / √x+h+1 + √x+1 f`(x)=lim x+h+1-x-1 / h(√x+h+1 + √x+1) f`(x)=lim 1 / h(√x+h+1 + √x+1) f`(x)= 1 / 2(√x+1) So that's f`(x). The problem is, I'm not sure where to go from here. I'm trying to find the equation of the line, and it has to be parallel to x-6y+4=0. Usually when I do a question like this, I'm given the point, but what do you do when you're given a line parallel to the tangent?