Question

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.

Answer 1

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.