It is the straight line that, when crossing the curved line, it is perpendicular to the curve.
The following figure shows the normal straight line to the curve $$\displaystyle y=\frac{1}{x-1}+1$$:
Two functions $$f(x),g(x)$$ will be normal in a point if, in the crossing point $$a$$, it is satisfied that:$$$f'(a)\cdot g'(a)=-1$$$
The following table shows several values of slopes of perpendicular straight lines:
| $$f'(a)$$ | $$g'(a)$$ |
| $$1$$ | $$-1$$ |
| $$2$$ | $$\displaystyle -\frac{1}{2}$$ |
| $$-3$$ | $$\displaystyle \frac{1}{3}$$ |
| $$\displaystyle \frac{3}{8}$$ | $$\displaystyle -\frac{8}{3}$$ |
The general expression of the normal straight line to $$f(x)$$ at a point $$a$$ is:$$$\displaystyle y-f(a)=-\frac{1}{f'(a)}\cdot (x-a)$$$
Solve the figure showed previously, that is, find the normal straight line to $$f(x)=\displaystyle \frac{1}{x-1}+1$$ at the point $$a=2$$:
a) The slope of the curve at the crossing point is:$$$\begin{array}{rcl} \displaystyle f'(x)& =& -\frac{1}{(x-1)^2} \\ f'(2)& = &-1\end{array}$$$And the slope of the straight line is: $$$\displaystyle m=-\frac{1}{f'(2)}=1$$$
b) Using the above mentioned, the straight line will go through $$$(a,f(a))=(2,2)$$$
Finally, the equation of the normal straight line is:$$$\begin{array}{rcl}y-2 & = & 1\cdot (x-2) \\ y & = & x \end{array}$$$Consistent with what we can observe in the figure.
Find the tangent straight line to the function $$y=\sqrt{x}$$ at the point $$x=0$$, as well as its normal straight line.
a) We start by looking at the derivative of the function at $$x=0$$.
However we can see that it does not exist. We then have to compute the limit as $$x$$ goes to zero from the right: $$$\displaystyle \begin{array}{l} y'(x)=\frac{1}{2\sqrt{x}} \\ \lim_{x \to 0} y'(x)=\lim_{x \to 0} \frac{1}{2\sqrt{x}}=\infty\end{array}$$$
b) Since the formula $$y=a\cdot x+b$$ is not useful when we have an infinite slope, we have to realize that it is the axis that is normal to curve. That is if we take the straight line $$x=0$$ we obtain a straight line with infinite slope.
c) Finally, we should realize that the line perpendicular to the straight line defined by the $$y$$ axis is the $$x$$ axis. That is, by the line defined by $$y=0$$.