question given $f(x)=-3x^{2}-2x$, find the equation of the tangent line of $f$ at the point where $x = 2$.
question given $f(x)=-3x^{2}-2x$, find the equation of the tangent line of $f$ at the point where $x = 2$.
Answer
Explanation:
Step1: Find the derivative of the function
The derivative of $f(x)=-3x^{2}-2x$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=-6x - 2$.
Step2: Find the slope of the tangent line at $x = 2$
Substitute $x = 2$ into $f'(x)$: $f'(2)=-6\times2-2=-12 - 2=-14$.
Step3: Find the y - coordinate of the point on the function at $x = 2$
Substitute $x = 2$ into $f(x)$: $f(2)=-3\times2^{2}-2\times2=-3\times4 - 4=-12 - 4=-16$.
Step4: Use the point - slope form to find the equation of the tangent line
The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(2,-16)$ and $m=-14$. So $y+16=-14(x - 2)$. Expand to get $y+16=-14x + 28$. Rearrange to the slope - intercept form $y=-14x+12$.
Answer:
$y=-14x + 12$