Explanation

(c)
As the curve crosses $y$-axis $ie,x = 0$
$\therefore\ \ \ \ \ y = 4e^{- 0} \Rightarrow y = 4$
Given, $y = 4e^{- \ \frac{x}{4}}$
$\Rightarrow \ \ \frac{dy}{dx} = 4e^{- \frac{x}{4}}\left( - \frac{1}{4} \right) = - e^{- \frac{x}{4}}$
$\Rightarrow \ \ \ \ \left( \frac{dy}{dx} \right)_{(0,4)} = - e^{- 0} = - 1$
$\therefore\ \ \$Equation of tangent at (0,4) is
$y - 4 = - 1(x - 0)$
$\Rightarrow \ \ \ \ \ \ x + y = 4$