find an equation for the tangent line to the graph of y=(x^3 - 9x)^12 at the point (3,0). the equation of…

Explanation:

Step1: Use chain - rule to find derivative

Let $u = x^{3}-9x$, then $y = u^{12}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=12u^{11}=12(x^{3}-9x)^{11}$, and $\frac{du}{dx}=3x^{2}-9$. So $\frac{dy}{dx}=12(x^{3}-9x)^{11}(3x^{2}-9)$.

Step2: Evaluate the derivative at $x = 3$

Substitute $x = 3$ into $\frac{dy}{dx}$: When $x = 3$, $x^{3}-9x=3^{3}-9\times3=27 - 27=0$. So $\frac{dy}{dx}\big|_{x = 3}=12\times0^{11}\times(3\times3^{2}-9)=0$.

Step3: Use point - slope form to find tangent line equation

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,0)$ and $m = 0$. $y-0=0\times(x - 3)$.

Answer:

$y = 0$