Hey guys, 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.

Question

shadowfiend · Answer

So to do this, you need to rewrite $x - 6y + 4$ into a form where you can easily find the slope ($y = mx + b$). Once you have the slope, you'll have to set up $y = mx + b$ again, with the $m$ you found above, and plugging in $\sqrt{x + 1}$ for $y$, so $\sqrt{x + 1} = mx + b$, with $m$ from the previous step, and solve for b.