Find an equation for the line that is tangent to the curve \( y=x^{3}-4 x \) at the point \( (2,0) \)

To find the equation of the tangent line to the curve \( y = x^3 - 4x \) at the point \( (2, 0) \), we need to follow these steps: 1. **Find the derivative of the function** to determine the slope of the tangent line at the given point. 2. **Evaluate the derivative at \( x = 2 \)** to find the slope at that point. 3. **Use the point-slope form of the equation of a line** to write the equation of the tangent line. ### Step 1: Find the derivative The function is given by: \[ y = x^3 - 4x \] To find the derivative \( y' \): \[ y' = \frac{d}{dx}(x^3) - \frac{d}{dx}(4x) = 3x^2 - 4 \] ### Step 2: Evaluate the derivative at \( x = 2 \) Now we evaluate the derivative at \( x = 2 \): \[ y'(2) = 3(2^2) - 4 = 3(4) - 4 = 12 - 4 = 8 \] So, the slope of the tangent line at the point \( (2, 0) \) is \( 8 \). ### Step 3: Use the point-slope form The point-slope form of the equation of a line is given by: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \( (x_1, y_1) \) is the point on the line. Here, \( m = 8 \) and \( (x_1, y_1) = (2, 0) \). Substituting these values into the point-slope form: \[ y - 0 = 8(x - 2) \] This simplifies to: \[ y = 8(x - 2) \] Expanding this gives: \[ y = 8x - 16 \] ### Final Answer The equation of the tangent line to the curve \( y = x^3 - 4x \) at the point \( (2, 0) \) is: \[ \boxed{y = 8x - 16} \]