To solve this problem, we first need to find the point of intersection \( C \) where the curve \( y = x^3 + 8 \) intersects the x-axis. At the x-axis, \( y = 0 \).
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Finding point C: \[ 0 = x^3 + 8 \] \[ x^3 = -8 \] \[ x = -2 \] Therefore, point \( C = (-2, 0) \).
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Finding the derivative of the curve: We need to find the slope of the tangent at point \( C \). \[ y' = \frac{d}{dx}(x^3 + 8) = 3x^2 \] Evaluating the derivative at \( x = -2 \): \[ y'(-2) = 3(-2)^2 = 3 \cdot 4 = 12 \] Thus, the slope of the tangent at point \( C \) is 12.
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Finding the equation of the tangent line at point C: The equation of the tangent line can be written in point-slope form: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency \( C(-2, 0) \). Substituting the values: \[ y - 0 = 12(x + 2) \] Simplifying this: \[ y = 12x + 24 \]
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Finding the y-intercept (point D) of the tangent line: The y-intercept occurs when \( x = 0 \): \[ y = 12(0) + 24 = 24 \] Thus, the coordinates of point \( D \) are \( (0, 24) \).
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Finding the distance \( CD \): We can find distance \( CD \) using the distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \( (x_1, y_1) = (-2, 0) \) and \( (x_2, y_2) = (0, 24) \). Substituting the coordinates: \[ CD = \sqrt{(0 - (-2))^2 + (24 - 0)^2} = \sqrt{(2)^2 + (24)^2} \] Calculating: \[ CD = \sqrt{4 + 576} = \sqrt{580} = \sqrt{4 \cdot 145} = 2 \sqrt{145} \]
Thus, the distance \( CD \) is \( \boxed{2\sqrt{145}} \).
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