given $f(x)=3x^{2}+2x + 2$, find the equation of the tangent line of $f$ at the point where $x=-4$.

Explanation:

Step1: Find the derivative of the function

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

Step2: Find the slope of the tangent line

Substitute $x = - 4$ into $f'(x)$. So $m=f'(-4)=6\times(-4)+2=-24 + 2=-22$.

Step3: Find the y - coordinate of the point

Substitute $x=-4$ into $f(x)$. $f(-4)=3\times(-4)^{2}+2\times(-4)+2=3\times16-8 + 2=48-8 + 2=42$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-4,42)$ and $m=-22$. So $y - 42=-22(x + 4)$.

Step5: Simplify the equation

Expand the right - hand side: $y-42=-22x-88$. Then $y=-22x-88 + 42=-22x-46$.

Answer:

$y=-22x - 46$