Problem. \( 4: \) Find the equation of the line tangent to \( f(x)=-x^{3}+9 x^{2}-6 x+9 \) at the point \( (7,65) \) \[ y=\square \]

To find the equation of the tangent line to the function \( f(x) = -x^3 + 9x^2 - 6x + 9 \) at the point \( (7, 65) \), we need to follow these steps: 1. **Find the derivative of \( f(x) \)** to determine the slope of the tangent line. 2. **Evaluate the derivative at \( x = 7 \)** 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 of \( f(x) \) The derivative \( f'(x) \) is calculated as follows: \[ f'(x) = \frac{d}{dx}(-x^3 + 9x^2 - 6x + 9) \] Using the power rule: \[ f'(x) = -3x^2 + 18x - 6 \] ### Step 2: Evaluate the derivative at \( x = 7 \) Now we substitute \( x = 7 \) into the derivative to find the slope at that point: \[ f'(7) = -3(7^2) + 18(7) - 6 \] Calculating \( 7^2 \): \[ 7^2 = 49 \] Now substituting: \[ f'(7) = -3(49) + 18(7) - 6 \] \[ = -147 + 126 - 6 \] \[ = -147 + 126 - 6 = -27 \] So, the slope of the tangent line at the point \( (7, 65) \) is \( -27 \). ### Step 3: Use the point-slope form of the equation of a line 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, \( m = -27 \), \( x_1 = 7 \), and \( y_1 = 65 \). Substituting these values into the point-slope form: \[ y - 65 = -27(x - 7) \] Now, we can simplify this equation: \[ y - 65 = -27x + 189 \] \[ y = -27x + 189 + 65 \] \[ y = -27x + 254 \] Thus, the equation of the tangent line is: \[ \boxed{y = -27x + 254} \]