question\ngiven $f(x)=3x^{2}-3x - 2$, find the equation of the tangent line of $f$ at the point where $x=-2$.

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=3x^{2}-3x - 2$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=6x-3$.

Step2: Find the slope of the tangent line

Substitute $x = - 2$ into $f'(x)$. So $m=f'(-2)=6\times(-2)-3=-12 - 3=-15$.

Step3: Find the y - coordinate of the point

Substitute $x=-2$ into $f(x)$. $f(-2)=3\times(-2)^{2}-3\times(-2)-2=3\times4 + 6-2=12 + 6-2=16$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-2,16)$ and $m=-15$. So $y - 16=-15(x+2)$.

Step5: Simplify the equation

$y-16=-15x-30$, then $y=-15x - 14$.

Answer:

$y=-15x - 14$