Find the coordinates of points where the graph of \( f(x)=-x^{3}+6 x^{2}-9 x+5 \) has horizontal tangents. \( (x, y)=(\square) \) (smaller \( x \)-value) \( (x, y)=(\square) \)

To find the points where the graph of the function \( f(x) = -x^3 + 6x^2 - 9x + 5 \) has horizontal tangents, we need to determine where the derivative \( f'(x) \) equals zero. First, compute the derivative: \[ f'(x) = -3x^2 + 12x - 9. \] Next, set the derivative equal to zero: \[ -3x^2 + 12x - 9 = 0. \] Dividing the entire equation by -3 yields: \[ x^2 - 4x + 3 = 0. \] Now, we can factor the quadratic: \[ (x - 1)(x - 3) = 0. \] This gives us \( x = 1 \) and \( x = 3 \). Now, substitute these \( x \)-values back into the original function \( f(x) \) to find the corresponding \( y \)-coordinates. For \( x = 1 \): \[ f(1) = -(1)^3 + 6(1)^2 - 9(1) + 5 = -1 + 6 - 9 + 5 = 1. \] For \( x = 3 \): \[ f(3) = -(3)^3 + 6(3)^2 - 9(3) + 5 = -27 + 54 - 27 + 5 = 5. \] Hence, the coordinates of the points where the graph has horizontal tangents are: \[ (x, y) = (1, 1) \quad \text{(smaller \( x \)-value)} \] \[ (x, y) = (3, 5). \]