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Imagine a car spinning out on an icy circular track — the exact straight trajectory it slides along is the tangent line to the curve at the point where it lost grip.
In calculus, finding the Cartesian equation of this line requires just two pieces of information: a single contact point (x0,y0)(x_0, y_0) and the instantaneous slope mm at that point.

The Tangent Recipe

The derivative f′(x)f'(x) gives the slope formula for any point on the curve, so evaluating f′(x0)f'(x_0) yields the numerical slope m=f′(x0)m = f'(x_0).
Once you have the slope mm and the point (x0,y0)(x_0, y_0) where y0=f(x0)y_0 = f(x_0), you substitute them directly into the point-slope form of a line: y−y0=m(x−x0)y - y_0 = m(x - x_0).
How do we put this formula into action on an actual cubic polynomial?

Worked Example: Polynomial Curve

Let's find the Cartesian equation of the tangent line to the curve f(x)=x3−4x+1f(x) = x^3 - 4x + 1 at the point where x=2x = 2.
First, find the yy-coordinate by plugging x0=2x_0 = 2 into the original function: y0=f(2)=(2)3−4(2)+1=1y_0 = f(2) = (2)^3 - 4(2) + 1 = 1, giving the contact point (2,1)(2, 1).
Next, compute the derivative using the power rule to get f′(x)=3x2−4f'(x) = 3x^2 - 4, and evaluate it at x=2x = 2 to find the slope m=f′(2)=3(2)2−4=8m = f'(2) = 3(2)^2 - 4 = 8.
Finally, substitute (x0,y0)=(2,1)(x_0, y_0) = (2, 1) and m=8m = 8 into the point-slope form:
y−1=8(x−2)  ⟹  y−1=8x−16  ⟹  y=8x−15y - 1 = 8(x - 2) \implies y - 1 = 8x - 16 \implies y = 8x - 15