find an equation for the line tangent to the graph of $f(x)=-2xe^{x}$ at the point $(a,f(a))$ for $a = 3$.

Explanation:

Step1: Find the derivative of (f(x))

Use the product - rule ((uv)^\prime = u^\prime v+uv^\prime), where (u=-2x) and (v = e^{x}). (u^\prime=-2) and (v^\prime=e^{x}), so (f^\prime(x)=-2e^{x}-2xe^{x}=-2e^{x}(1 + x)).

Step2: Evaluate (f(3))

Substitute (x = 3) into (f(x)=-2xe^{x}), then (f(3)=-2\times3\times e^{3}=-6e^{3}).

Step3: Evaluate (f^\prime(3))

Substitute (x = 3) into (f^\prime(x)=-2e^{x}(1 + x)), then (f^\prime(3)=-2e^{3}(1 + 3)=-8e^{3}).

Step4: Use the point - slope form (y - y_{1}=m(x - x_{1}))

Here (x_{1}=3), (y_{1}=-6e^{3}) and (m=-8e^{3}). So (y+6e^{3}=-8e^{3}(x - 3)).

Step5: Simplify the equation

[ \begin{align*} y+6e^{3}&=-8e^{3}x+24e^{3}\ y&=-8e^{3}x + 18e^{3} \end{align*} ]

Answer:

(y=-8e^{3}x + 18e^{3})