Questions: Find an equation for the tangent line to the graph of y=(x^3-4x)^10 at the point (-2,0). The equation of the tangent line is y=

Find an equation for the tangent line to the graph of y=(x^3-4x)^10 at the point (-2,0).

The equation of the tangent line is y=
Transcript text: Find an equation for the tangent line to the graph of $y=\left(x^{3}-4 x\right)^{10}$ at the point $(-2,0)$. The equation of the tangent line is $y=$ $\square$ (Simplify your answer.)
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Solution

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Solution Steps

Step 1: Define the Function and Compute its Derivative

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: \[ y' = \frac{d}{dx} \left( (x^3 - 4x)^{10} \right) = 10(x^3 - 4x)^9 \cdot (3x^2 - 4) \]

Step 2: Evaluate the Derivative at the Given Point

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 \( 0 \).

Step 3: Write the Equation of the Tangent Line

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: \[ y = 0 \]

Final Answer

\(\boxed{y = 0}\)

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