The tangent line to $f$ at $x = a$ passes through the point $(a, f(a))$ with slope $f'(a)$. Its equation is $y = f(a) + f'(a)(x - a)$, read aloud: y equals f of a, plus f prime of a times x minus a. Here $a$ is the x-value where the line touches, $f(a)$ is the height there, $f'(a)$ is the slope there, a single number, and $x$ and $y$ are the coordinates of any point on the line. It is the point-slope form of a line, $y = y_1 + m(x - x_1)$, with the point $(x_1, y_1) = (a, f(a))$ and $m = f'(a)$. The trap: the slope is the NUMBER $f'(a)$, the derivative evaluated at $a$. Leaving $f'(x)$ in the equation gives a curve, not a line.

For $f(x) = x^3 - 3x$, $f'(x) = 3x^2 - 3$. At $a = 2$: $f(2) = 8 - 6 = 2$ and $f'(2) = 12 - 3 = 9$, so the tangent is $y = 2 + 9(x - 2)$, which tidies to $y = 9x - 16$. At $a = 1$: $f'(1) = 0$, so the tangent is the horizontal line $y = -2$, and the curve turns there. Below, slide the point to $a = 2$ and untick the secant: the read-out writes the tangent's equation from $f(a)$ and $f'(a)$, and dragging $a$ shows the line tilting with the curve.