Let's solve the problem step-by-step: **Given:** \( f(x) = x^2 + \frac{2}{x} - 10 \) Find the equation of the tangent line at \( x = 8 \). --- ### **Step 1: Find the point on the curve at \( x = 8 \)** \[ f(8) = (8)^2 + \frac{2}{8} - 10 = 64 + 0.25 - 10 = 54.25 \] So the point is \( (8, 54.25) \). --- ### **Step 2: Find the derivative \( f'(x) \)** \[ f(x) = x^2 + 2x^{-1} - 10 \] \[ f'(x) = 2x - 2x^{-2} \] \[ = 2x - \frac{2}{x^2} \] --- ### **Step 3: Find the slope at \( x = 8 \)** \[ f'(8) = 2(8) - \frac{2}{8^2} = 16 - \frac{2}{64} = 16 - 0.03125 = 15.96875 \] --- ### **Step 4: Write the equation of the tangent line** The equation of a tangent line at \( (x_0, y_0) \) with slope \( m \) is: \[ y - y_0 = m(x - x_0) \] Plug in the values: - \( x_0 = 8 \) - \( y_0 = 54.25 \) - \( m = 15.96875 \) \[ y - 54.25 = 15.96875(x - 8) \] \[ y = 15.96875(x - 8) + 54.25 \] --- ### **Final Answer** \[ \boxed{y = 15.96875(x - 8) + 54.25} \] Or simplified: \[ \boxed{y = 15.96875x - 73.5} \]

Certainly! Here are two lines of pure theory related to the problem: 1. **The equation of the tangent line to a curve at a given point is derived using the point-slope form, where the slope is the derivative of the function evaluated at that point.** 2. **The derivative of a function provides the rate of change (slope) of the tangent line at any point on the curve, enabling the construction of the tangent line's equation at that point.**