Find the equation of the tangent line and normal line to the curve e^(-x)-y=0 at (-1,0).

Question

Find the equation of the tangent line and normal line to the curve e^(-x)-y=0 at (-1,0).

Answer 1

Let (x₀, y₀) is tangent point. Then y₀ = e^(-x₀); y' = - e^-x; y'(x₀) = - e^(- x₀). Equation of tangent line:

y = e^(- x₀) - e^(- x₀)(x - x₀). This line passes point (-1, 0). So, 0 = e^(- x₀) - e^(- x₀)(- 1 - x₀);

Because e^(-x₀) ≠ 0 we have 1 + 1 + x₀ = 0; x₀ = - 2; y₀ = e² and y^'(- 2) = - e².

Equation of tangent line y = e² - e²(x + 2) = - e²(x + 1);

Equation of normal line y = e² + e^-2(x + 2)