How is my textbook using this point slope equation of a line?

Question

I'm learning about Newton's Method via this text:

How is the point slope method being used? Why are we subbing $x_2$ and 0 in the wrong places in this equation:

$$0 - f(x_1) = f'(x_1)(x_2 - x_1)$$

Is the author just subbing in a point for the point-slope equation of a line? Because from what I see, the point slope form of a line uses the slope and plugs in the given point in the 2nd positions:

https://www.mathsisfun.com/algebra/line-equation-point-slope.html

Answer 1

If $(x_a, y_a) $ and a point on a line and the slope is $m$ the line formula is

$y - y_a = m(x - x_a)$.

So here $m = f'(x_1)$ and $(x_1, f(x_1))$ is one point on the line so

$y - f(x_1) = f'(x_1)(x - x_1)$

is the equation of the line.

So if $(w, v)$ is a point on the line, it's satisfy the equation we'd get if we plugged in $y = v$ and $x = w$ and get

$v - f(x_1) = f'(x_1)(w - x_1)$

Now we want to plug in the $x$ intercept $(x_2, 0)$.

So $y = 0$ and $x = x_2$ and we plug them in:

So $0 - f(x_1) = f'(x_1)(x_2 - x_1)$.

Answer 2

The slope at the point $x_1$ is

$$ f'(x_1) = \left.\frac{{\rm d}f}{{\rm d}x}\right|_{x=x_1} $$

that means that the equation of the line with slope $f'(x_1)$ going through the point $(x_1, y_1)$ is

$$ y - y_1 = f'(x_1)(x-x_1) $$

You also want to find the point $x_2$ that makes $y=0$. Substituting this constraint in the equation above you get

$$ 0-y_1 = f'(x_1)(x_2 - x_1) $$

or equivalently

$$ x_2 = x_1 - \frac{y_1}{f'(x_1)} $$

Answer 3

Note that to find the $x$-intercept we let $y=0$

The equation $$0 - f(x_1) = f'(x_1)(x_2 - x_1)$$ is result of replacing $y$ with $0$ in order to find the x-intercept of the tangent line.

They replaced $ y$ with $0$ and solved for $ x_2$.