We are given the function \( y = (x^3 - 4x)^{10} \). To find the equation of the tangent line at the point \((-2, 0)\), we first need to compute the derivative of the function, \( y' \), which represents the slope of the tangent line at any point \( x \).
The derivative of the function is:
\[
y' = \frac{d}{dx} \left( (x^3 - 4x)^{10} \right) = 10(x^3 - 4x)^9 \cdot (3x^2 - 4)
\]
Next, we evaluate the derivative at \( x = -2 \) to find the slope of the tangent line at the point \((-2, 0)\).
Substituting \( x = -2 \) into the derivative:
\[
y'(-2) = 10((-2)^3 - 4(-2))^9 \cdot (3(-2)^2 - 4) = 10(0)^9 \cdot 8 = 0
\]
The slope of the tangent line at \((-2, 0)\) is \( 0 \).
The equation of a tangent line at a point \((x_1, y_1)\) with slope \( m \) is given by:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( m = 0 \), \( x_1 = -2 \), and \( y_1 = 0 \):
\[
y - 0 = 0(x + 2)
\]
Simplifying, we find:
\[
y = 0
\]