Question

find an equation of the tangent line to the graph of f at the given point. y = x^3 - 3x + 1, (2, 3) y =

Explanation:

Step1: Find the derivative of the function

The derivative of $y = x^{3}-3x + 1$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=3x^{2}-3$.

Step2: Evaluate the derivative at the given x - value

Substitute $x = 2$ into $y'$. So $y'(2)=3\times2^{2}-3=3\times4 - 3=12 - 3 = 9$. The value of the derivative at $x = 2$ is the slope $m$ of the tangent line, so $m = 9$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(2,3)$ and $m = 9$. Substitute these values: $y - 3=9(x - 2)$.

Step4: Simplify the equation

Expand the right - hand side: $y-3=9x-18$. Then add 3 to both sides to get $y=9x - 15$.

Answer:

$y = 9x-15$