# 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?

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The given line is not the normal line. The normal to $y=f(x)$ at the point $P=(a,f(a))$ is the line through $P$ that is perpendicular to the tangent line at $P$. It will turn out, once you have found $P$, that the given line does not pass through $P$. –  André Nicolas Oct 4 '11 at 5:41
In fact, the answer would be the same for any value of 4. –  TonyK Nov 21 '13 at 15:50

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.

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Should I reduce my answer for the possibility of this being homework? –  Altar Ego Oct 4 '11 at 5:30
Let's not worry about it, answer is fine, particularly since the question does not have a homework label. As a minor point, the $a$ that I used actually has a purpose. I have seen all too many times students use $x$ in roughly similar situations, and then end up with a non-linear "equation" for the tangent line. –  André Nicolas Oct 4 '11 at 5:34
That is a helpful approach. –  Altar Ego Oct 4 '11 at 5:35

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.

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For a while, I will try to upvote any answer that mentions drawing a diagram. So often crucially important! –  André Nicolas Oct 4 '11 at 5:48

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?

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i got the perpendicular slope as 1/2. i got f'(x)=1/2(x-3)^-1/2 = 1/2. is that the same? is that wrong? –  user17108 Oct 4 '11 at 5:34
from there i did: 1/2(√(x-3)/x-3) = 1/2. not sure if that is right at this point. –  user17108 Oct 4 '11 at 5:39
You are complicating things. $\frac{1}{2\sqrt{x-3}}=\frac{1}{2}$. Therefore $2\sqrt{x-3}=2$. Continue. –  André Nicolas Oct 4 '11 at 5:45
I am more than open to learning how to do this a less complicated way. I was taught to change √(x-3) to (x-3)^-1/2. I know this is just algebra, but can you illustrate how you got to 2√(x-3)=2 from your f'? I think this is my main issue. I'm not sure with how to do this derivative and am coming up with 2 x's at the end. –  user17108 Oct 4 '11 at 5:56
I just flipped both of them over (OK, in fancier language, found the reciprocal of both sides). Or you can "cross-multiply." Rationalizing the denominator can be helpful occasionally, but in this case it complicates things. Or you could multiply both sides by $2$, get $\frac{1}{\sqrt{x-3}}=1$ and then what to do will be even clearer. –  André Nicolas Oct 4 '11 at 6:01