# Math Insight

### Tangent and normal lines

One fundamental interpretation of the derivative of a function is that it is the slope of the tangent line to the graph of the function. (Still, it is important to realize that this is not the definition of the thing, and that there are other possible and important interpretations as well).

The precise statement of this fundamental idea is as follows. Let $f$ be a function. For each fixed value $x_o$ of the input to $f$, the value $f'(x_o)$ of the derivative $f'$ of $f$ evaluated at $x_o$ is the slope of the tangent line to the graph of $f$ at the particular point $(x_o,f(x_o))$ on the graph.

Recall the point-slope form of a line with slope $m$ through a point $(x_o,y_o)$: $$y-y_o=m(x-x_o)$$ In the present context, the slope is $f'(x_o)$ and the point is $(x_o,f(x_o))$, so the equation of the tangent line to the graph of $f$ at $(x_o,f(x_o))$ is $$y-f(x_o)=f'(x_o)(x-x_o)$$

The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent. A person might remember from analytic geometry that the slope of any line perpendicular to a line with slope $m$ is the negative reciprocal $-1/m$. Thus, just changing this aspect of the equation for the tangent line, we can say generally that the equation of the normal line to the graph of $f$ at $(x_o,f(x_o))$ is $$y-f(x_o)={-1\over f'(x_o)}(x-x_o).$$

The main conceptual hazard is to mistakenly name the fixed point ‘$x$’, as well as naming the variable coordinate on the tangent line ‘$x$’. This causes a person to write down some equation which, whatever it may be, is not the equation of a line at all.

Another popular boo-boo is to forget the subtraction $-f(x_o)$ on the left hand side. Don't do it.

So, as the simplest example: let's write the equation for the tangent line to the curve $y=x^2$ at the point where $x=3$. The derivative of the function is $y'=2x$, which has value $2\cdot 3=6$ when $x=3$. And the value of the function is $3\cdot 3=9$ when $x=3$. Thus, the tangent line at that point is $$y-9=6(x-3)$$ The normal line at the point where $x=3$ is $$y-9={-1\over 6}(x-3)$$

So the question of finding the tangent and normal lines at various points of the graph of a function is just a combination of the two processes: computing the derivative at the point in question, and invoking the point-slope form of the equation for a straight line.

#### Exercises

1. Write the equation for both the tangent line and normal line to the curve $y=3x^2-x+1$ at the point where $x=1$.
2. Write the equation for both the tangent line and normal line to the curve $y=(x-1)/(x+1)$ at the point where $x=0$.