The equation of the tangent to the curve $y=1-e^{x / 2}$ at the point of intersection with the y-axis, is
Answer & explanation
Correct answer: option 1
We have,
$y=1-e^{x / 2} \Rightarrow \frac{d y}{d x}=-\frac{1}{2} e^{x / 2}$
The curve $y=1-e^{x / 2}$ meets y-axis at (0, 0)
∴ $\left(\frac{d y}{d x}\right)_{(0,0)}=-\frac{1}{2}$
The equation of the tangent at (0, 0) is
$y-0=-\frac{1}{2}(x-0) \Rightarrow x+2 y=0$