### 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

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

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