To find the equation of the normal to the curve \( y = x^3 - 3x + 2 \) at the point \( (2,4) \)
Step 1: Compute the Derivative
The derivative of the given function represents the slope of the tangent line: \[ \frac{dy}{dx} = \frac{d}{dx} (x^3 - 3x + 2) \] \[ \frac{dy}{dx} = 3x^2 - 3 \] Step 2: Find the Slope at \( (2,4) \)
Substituting \( x = 2 \): \[ \frac{dy}{dx} \bigg|_{x=2} = 3(2)^2 - 3 = 3(4) - 3 = 12 - 3 = 9 \] Thus, the slope of the tangent at \( (2,4) \) is 9.
Step 3: Compute the Slope of the Normal
The slope of the normal line is the negative reciprocal of the tangent's slope: \[ m_{\text{normal}} = -\frac{1}{9} \] Step 4: Equation of the Normal Line
Using the point-slope form of the equation of a line: \[ y - y_1 = m (x - x_1) \] Substituting \( (x_1, y_1) = (2,4) \) and \( m = -\frac{1}{9} \): \[ y - 4 = -\frac{1}{9} (x - 2) \] Multiplying both sides by 9 to eliminate fractions: \[ 9(y - 4) = -(x - 2) \] \[ 9y - 36 = -x + 2 \] \[ x + 9y = 38 \] Final Answer: \( \mathbf{x + 9y = 38} \)