Find the equation of the tangent to the curve | vedclass

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(A) The given curve is $y = 1 - e^{x/2}$.To find the point where the curve intersects the $y$-axis,we set $x = 0$.Substituting $x = 0$ into the equati

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$x + 2y = 0$

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(A) The given curve is $y = 1 - e^{x/2}$.To find the point where the curve intersects the $y$-axis,we set $x = 0$.Substituting $x = 0$ into the equation: $y = 1 - e^{0/2} = 1 - 1 = 0$.So,the point of intersection is $(0, 0)$.Now,find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent:$\frac{dy}{dx} = \frac{d}{dx}(1 - e^{x/2}) = -e^{x/2} \cdot \frac{1}{2} = -\frac{1}{2}e^{x/2}$.The slope of the tangent at $(0, 0)$ is $m = \left(\frac{dy}{dx}\right)_{(0,0)} = -\frac{1}{2}e^{0} = -\frac{1}{2}$.The equation of the tangent line passing through $(x_1, y_1) = (0, 0)$ with slope $m = -\frac{1}{2}$ is given by $y - y_1 = m(x - x_1)$.$y - 0 = -\frac{1}{2}(x - 0)$.$y = -\frac{1}{2}x$.$2y = -x$,which simplifies to $x + 2y = 0$.

Find the equation of the tangent to the curve $y = 1 - e^{x/2}$ at the point where it intersects the $y$-axis.

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