Find an equation of the tangent line to the graph of the function at the given point. \[ h(x)=-x e^{5-x}, \quad(5,-5) \]

To find the equation of the tangent line to the function \( h(x) = -x e^{5-x} \) at the point \( (5, -5) \), we need to determine the slope of the tangent line at that point. This requires calculating the derivative \( h'(x) \) and then evaluating it at \( x = 5 \). First, we find the derivative \( h'(x) \) using the product rule: \[ h'(x) = \frac{d}{dx}(-x) \cdot e^{5-x} + (-x) \cdot \frac{d}{dx}(e^{5-x}). \] Calculating each part, we have: \[ \frac{d}{dx}(-x) = -1, \] and using the chain rule for \( e^{5-x} \): \[ \frac{d}{dx}(e^{5-x}) = e^{5-x} \cdot (-1) = -e^{5-x}. \] Substituting these back into the product rule gives us: \[ h'(x) = -e^{5-x} + x e^{5-x} = (x - 1)e^{5-x}. \] Next, we evaluate \( h'(x) \) at \( x = 5 \): \[ h'(5) = (5 - 1)e^{5-5} = 4 \cdot 1 = 4. \] Now, with the slope \( m = 4 \) and the point \( (5, -5) \), we can use the point-slope form of a line: \[ y - y_1 = m(x - x_1), \] where \( (x_1, y_1) = (5, -5) \): \[ y - (-5) = 4(x - 5). \] Simplifying this gives: \[ y + 5 = 4x - 20 \quad \Rightarrow \quad y = 4x - 25. \] Thus, the equation of the tangent line to the graph of the function at the point \( (5, -5) \) is: \[ \boxed{y = 4x - 25}. \]