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If you zoom in close enough on any smooth curve, it looks almost entirely like a straight line. What if we use that simple line to hunt down impossible roots that no algebra formula can solve directly?

The Tangent Line as a Stepping Stone

A root is any input value xx where a function crosses zero, meaning f(x)=0f(x) = 0. When an exact root is out of reach, we make an initial guess x0x_0 and draw the tangent line to the curve at that point.
The tangent line acts as a linear approximation of the curve. Where the tangent line hits the x-axis becomes our next, much closer guess: xn+1x_{n+1}.
How do we write an exact formula to calculate where this tangent line hits the x-axis?

Deriving the Formula

Every line with slope mm passing through a point (xn,yn)(x_n, y_n) follows the point-slope equation: y−yn=m(x−xn)y - y_n = m(x - x_n). Since the slope of a curve at xnx_n is given by the derivative f′(xn)f'(x_n), the tangent line at (xn,f(xn))(x_n, f(x_n)) is:
y−f(xn)=f′(xn)(x−xn)y - f(x_n) = f'(x_n)(x - x_n)
To find the new x-intercept xn+1x_{n+1}, we set y=0y = 0 and replace xx with xn+1x_{n+1}:
0−f(xn)=f′(xn)(xn+1−xn)0 - f(x_n) = f'(x_n)(x_{n+1} - x_n)
Dividing both sides by f′(xn)f'(x_n) and adding xnx_n gives the Newton-Raphson formula: xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}