Tangent line to $$$y = x^{3} - 3 x + 2$$$ at $$$x = 2$$$

Calculate the tangent line to $$$y = x^{3} - 3 x + 2$$$ at $$$x = 2$$$.

We are given that $$$f{\left(x \right)} = x^{3} - 3 x + 2$$$ and $$$x_{0} = 2$$$.

Find the value of the function at the given point: $$$y_{0} = f{\left(2 \right)} = 4$$$.

The slope of the tangent line at $$$x = x_{0}$$$ is the derivative of the function, evaluated at $$$x = x_{0}$$$: $$$M{\left(x_{0} \right)} = f^{\prime }\left(x_{0}\right)$$$.

Find the derivative: $$$f^{\prime }\left(x\right) = \left(x^{3} - 3 x + 2\right)^{\prime } = 3 x^{2} - 3$$$ (for steps, see derivative calculator).

Hence, $$$M{\left(x_{0} \right)} = f^{\prime }\left(x_{0}\right) = 3 x_{0}^{2} - 3$$$.

Next, find the slope at the given point.

$$$m = M{\left(2 \right)} = 9$$$

Finally, the equation of the tangent line is $$$y - y_{0} = m \left(x - x_{0}\right)$$$.

Plugging the found values, we get that $$$y - 4 = 9 \left(x - 2\right)$$$.

Or, more simply: $$$y = 9 x - 14$$$.

The equation of the tangent line is $$$y = 9 x - 14$$$A.