Equation of the Tangent Line? (Derivatives)

[Velo]

So, I'm re-learning the derivatives right now, and I don't get how the Equation of the Tangent Line is something like

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

I know that the Equation of a Line should be something like y = mx + b, so I get how f(a) = b (since it's a pont in that line), and f'(a) = m (since f'(a) is the slope) in that equation... I don't understand why the x is replaced by (x-a) though?

 

[Deleted member 4993]

So, I'm re-learning the derivatives right now, and I don't get how the Equation of the Tangent Line is something like y = f(a) + f'(a)(x-a)

I know that the Equation of a Line should be something like y = mx + b, so I get how f(a) = b (since it's a pont in that line), and f'(a) = m (since f'(a) is the slope) in that equation... I don't understand why the x is replaced by (x-a) though?

In y = f(a) + (x-a)*f'(x) equation, f(a) is NOT, equal to "b"(or y-intercept).

b = f(a) - a*f'(a)

In this case, you have at x =a → y = f(a) [as it should be].

 

[tkhunny]

It might make more sense in this form...

$\displaystyle f(x) - f(a) = f'(x)(x-a)$

If the only equation of a line you can think of is y = mx + b, perhaps you should back up a bit and focus on a little more algebra.

 

[mmm4444bot]

The linear equation shown in post #3 is known as Point-Slope form.

We use this form, to write the equation of a line, when we know both the slope and the coordinates of a point on the line.

slope: f'(x)

x-coordinate of known point: a

y-coordinate of known point: f(a)

 

[gregk]

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

Just to clarify a little more, your equation is not yet in slope-intercept form, $\displaystyle y = mx + b$. But if you want it in that form, then you can do it by distributing the $\displaystyle f'(a)$ as follows:

\(\displaystyle \begin{align*}
y = f(a) + f'(a)(x - a) \quad&\Rightarrow\quad y = f(a) + f'(a)\cdot x - f'(a)\cdot a \\[5pt]
&\Rightarrow\quad y = xf'(a) + [f(a) - af'(a)].
\end{align*}\)

Now it fits the pattern: your slope is $\displaystyle m = f'(a)$ and your y-intercept is $\displaystyle b = f(a) - af'(a)$.

Of course, there's nothing particularly special about the slope-intercept form other than it being most commonly taught; you should be able to recognize a linear equation in any form and find its slope and intercepts using basic algebra.