Find an equation of the tangent line to the graph of $y = \sqrt{x-3}$ that is perpendicular to $6x+3y−4=0.$ 
Find an equation of the tangent line to the graph of $y= \sqrt{x-3}$ that is perpendicular to $6x+3y-4=0$. 

I don't understand what it's asking. Is this the normal line? How do I solve this?
 A: Since this may be homework (please correct me if it isn't), I will give a few hints only.
Hint 1: What is the slope of the given line $6x+3y-4=0$?
Hint 2: What is the slope of any line perpendicular to the given line?
Hint 3: So the "mystery" tangent line must have the slope reached in Hint 2.
Hint 4: Let $\ell$ be a tangent line to $y=\sqrt{x-3}$ at the point $(a,\sqrt{a-3})$.  In terms of $a$, what is the slope of $\ell$?
Hint 5: Combine the results obtained in the two previous hints. The rest should be familiar.   
A: Andre's answer is good. Another approach which may stand you in good stead beyond this particular question is: draw a diagram. Sketch the graph of $y=\sqrt{x-3}$ (it doesn't have to be a real good sketch, actually it's probably good enough just to draw some random curve), sketch the line $6x+3y-4=0$ (again, probably any line will do, if all we want is to work out what the question is asking). Draw any one of the many tangents to the graph of $y=\sqrt{x-3}$. Does the tangent you have just drawn meet the line $6x+3y-4=0$ at right angles? Probably not. Draw a different tangent to the graph of $y=\sqrt{x-3}$. Is this one perpendicular to the line $6x+3y-4=0$? Are you getting a feel for what the question is asking? 
Often, drawing a simple diagram not only helps you understand what a question is asking, it helps you see how to answer it. 
A: First, determine the slope of the line $6x + 3y - 4 = 0$. Here, $m = -2$.
Then we calculate the perpendicular slope to $-2$ as $1/2$ (why?).
Then we want to find where the slope of the tangent to $y = \sqrt{x - 3}$ is equal to $1/2$.
In other words, where $y' = \frac{1}{2\sqrt{x - 3}} = 1/2$. 
Can you take it from here?
