Question

The equation of the tangent to the curve $y=1-e^{\frac{x}{2}}$ at the point of intersection with the y-axis is

• A. x + 2y = 0
• B. 2x + y = 0
• C. x – y = 2
• D. none of these

Answer 1

The correct option is A x + 2y = 0

For y-axis, x = 0
$\therefore y=1-e^{\circ }=-1=0\therefore \frac{dy}{dx}=0-\frac{1}{2}{e}^{\frac{x}{2}},(\frac{dy}{dx}{)}_{(0,0)}=-\frac{1}{2}$
$\therefore$ Equation of tangent
$y-0=-\frac{1}{2}(x-0)\Rightarrow x+2y=0$