## Step 1: Equation of the Tangent Line Let the point of tangency be at \( (a, e^{-a}) \). The slope of the tangent to \( y = e^{-x} \) at \( x = a \) is: \[ y' = -e^{-x} \implies \text{slope at } x=a \text{ is } -e^{-a} \] So, the equation of the tangent at \( x = a \) is: \[ y - e^{-a} = -e^{-a}(x - a) \] \[ y = -e^{-a}(x - a) + e^{-a} \] \[ y = -e^{-a}x + a e^{-a} + e^{-a} \] \[ y = -e^{-a}x + e^{-a}(a + 1) \] **Explanation:** This step formulates the tangent line’s equation in terms of \( a \), which is the \( x \)-coordinate where the tangent touches the curve. --- ## Step 2: Find the Intercepts (Triangle Vertices) - **\( x \)-intercept:** Set \( y = \) \[ = -e^{-a}x + e^{-a}(a + 1) \implies x = a + 1 \] - **\( y \)-intercept:** Set \( x = \) \[ y = -e^{-a}() + e^{-a}(a + 1) = e^{-a}(a + 1) \] The triangle is bounded by the \( x \)-axis, \( y \)-axis, and the tangent. The base is the segment on the \( x \)-axis from \( x= \) to \( x=a+1 \), and the height is the \( y \)-intercept. The area \( A \) of the triangle is: \[ A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} (a + 1) \cdot [e^{-a}(a + 1)] = \frac{1}{2} (a + 1)^2 e^{-a} \] **Explanation:** This step finds the vertices of the triangle and expresses the area as a function of \( a \). --- ## **Final Answer** The area \( A \) as a function of \( a \) is: \[ \boxed{A(a) = \frac{1}{2} (a + 1)^2 e^{-a}} \] where \( a > \) (since the triangle is in the first quadrant). To maximize the area, one would find the absolute maximum of \( A(a) \) over \( a > \).