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Let $f:A \rightarrow \mathbb R$ be a differentiable function, where $A$ is an open interval of real numbers.

Let $a \in A$.

Then the tangent of $f$ at $a$ is defined by the equation $f(a)+f'(a)(x-a)$.

I understand the intuition behind choosing a point $f(a)$ through which the tangent will definitely pass through and another point $f'(a)$ - which, since it defines the slope of $f$ at $a$, the tangent line will surely pass through too.

Now, I don't understand the intuition behind why $f'(a)$ is multiplied by $(x-a)$ and not just $x$ - if we multiply $f'(a)$ by every real number, in a space with Cartesian coordinates we are going to have the range of the function be every point in the line which passes through $f(a)$ and $f'(a)$.

Why don't we define the tangent of $f$ at $a$ to be simply $f(a)+f'(a)x$?

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  • $\begingroup$ Let $g(x)$ be the function of the tangent line at $x=a$. One must have $g(a)=f(a)$. $\endgroup$ – Jack Mar 1 '17 at 21:31
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    $\begingroup$ Consider the case $f(x)=x+1$. What do you think the tangent line at $x=1$ should be? $\endgroup$ – Jack Mar 1 '17 at 21:33
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Because it doesn't work that way.

To start with, $f(a)$ is not a point. It's only a number. An example of a point on the graph of $f$ is $(a, f(a))$--two coordinates, $a$ and $f(a),$ to plot a point in the plane.

Next, $f'(a)$ is the slope of the tangent line. You may have realized that already, but remember that the equation of a line in the $x,y$ plane satisfies $y = b + mx$ if and only if $m$ is the slope and $b$ is the $y$-intercept of the line. So if we can figure out a number $b$ such that the tangent line at $(a, f(a))$ goes through the point $(0,b),$ then we can write $y = b + (f'(a))x.$

There is a more generally-applicable form of the equation of (almost) any line in the plane, which lets you write the equation for a line easily once you know its slope and any point that it passes through (not necessarily where it intersects the $y$ axis). For a line with slope $m$ passing through the point $(x_1,y_1),$ the equation is $$ y = y_1 + m(x - x_1). \tag1 $$ The equation can also be written in a slightly more symmetric form, $$ y - y_1 = m(x - x_1), $$ which tells us this is the equation of the line we get if we translate the line $y=mx$ upward $y_1$ units and $x_1$ units to the right so that it goes through the point $(x_1,y_1).$

For the tangent line, instead of an arbitrary point named $(x_1,y_1)$ we have one named $(a,f(a)),$ and instead of an arbitrary slope $m$ we have slope $f'(a).$ So where do these new symbols fit into Equation $1,$ and what does it look like after we update it to describe the tangent line?


Of course you still can write an equation for the tangent line in intercept-slope form if you really want to. Just observe that the equation of the tangent line says $$ y = f(a)+f'(a)(x-a) = \left( f(a) - af'(a) \right) + (f'(a))x, $$ so the $y$-intercept of the tangent line is $f(a) - af'(a).$

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  • $\begingroup$ A fundamental remark is that the equation of the tangent $y=f(a)+f'(a)(x-a)$ is the beginning ot Taylor expansion of $f$ around pooint a, i.e., $f(x)=f(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2+...$. In other words, it is the best affine approximation of function $f$ , up to 2nd degree terms... $\endgroup$ – Jean Marie Mar 1 '17 at 22:53
  • $\begingroup$ @JeanMarie That's very true, but I designed this answer to be read by someone who is just starting to learn about derivatives and their relationship to the tangent lines of graphs of functions. I don't expect the reader to know about Taylor expansions yet. ... Maybe I can add something at the end about "things you will see later." $\endgroup$ – David K Mar 1 '17 at 23:38
  • $\begingroup$ @I agree with you : I appreciate your understanding of the level at which the answers should be given. $\endgroup$ – Jean Marie Mar 2 '17 at 7:07
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What is the equation of the line passing through $(a,b)$ of slope $m?$ It is $y = b + m(x-a).$ So then, what is the equation of the line passing through $(a,f(a))$ of slope $f'(a)?$ It is

$$y = f(a) + f'(a)(x-a).$$

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