Question

equation of the tangent to the curve y = 1-e^-x/2 at the point where the curve cuts y axis is

Answer 1

$\textbf{Given:}$

$y=1-e^\frac{-x}{2}$

$\text{At the point where the tangent cuts y-axis, x=0}$

$\implies\,y=1-e^0=1-1=0$

$\text{Now,}$

$\frac{dy}{dx}=e^\frac{-x}{2}\,(\frac{-1}{2})$

$\text{Slope of tangent,}\;m=\frac{dy}{dx}_{(0,0)}=e^0\,(\frac{-1}{2})}$

$\implies\,m=\frac{-1}{2}$

$\text{The equation of tangent is}$

$y-y_1=m(x-x_1)$

$y-0=\frac{-1}{2}(x-0)$

$y=\frac{-x}{2}$

$2y=-x$

$\implies\boxed{\bf\,x+2y=0}$