Let's solve the problem: The function is \( g(x) = x^3 - 6x^2 + 9x + k \) and it is tangent to the x-axis. For the function to be tangent to the x-axis at some point \( x = a \): 1. \( g(a) = 0 \) (it touches the x-axis) 2. \( g'(a) = 0 \) (slope is zero, i.e., tangent) Let's find \( a \): First, compute the derivative: \[ g'(x) = 3x^2 - 12x + 9 \] Set \( g'(x) = 0 \): \[ 3x^2 - 12x + 9 = 0 \\ x^2 - 4x + 3 = 0 \\ (x-1)(x-3) = 0 \\ x = 1 \text{ or } x = 3 \] Now check \( g(x) = 0 \) for these values: For \( x = 1 \): \[ g(1) = 1^3 - 6(1)^2 + 9(1) + k = 1 - 6 + 9 + k = 4 + k \] Set \( g(1) = 0 \Rightarrow 4 + k = 0 \Rightarrow k = -4 \) For \( x = 3 \): \[ g(3) = 3^3 - 6(3^2) + 9(3) + k = 27 - 54 + 27 + k = 0 + k = k \] Set \( g(3) = 0 \Rightarrow k = 0 \) Let's check the options: 1. 1 2. -2 3. 0 **<--- This matches \( k = 0 \)** 4. -1 **Final Answer:** The value of \( k \) is **0**. Select option **0**.