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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 $P$.2011-10-04
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    In fact, the answer would be the same for any value of 4.2013-11-21

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