The tangent drawn at the point (0, 1) on the | vedclass

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(B) Given the curve $y = e^{2x}$.First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 2e^{2x}$.At the point $(0, 1)$,the

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$(-1/2, 0)$

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(B) Given the curve $y = e^{2x}$.First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 2e^{2x}$.At the point $(0, 1)$,the slope $m$ is: $m = \left. \frac{dy}{dx} \right|_{(0, 1)} = 2e^{2(0)} = 2(1) = 2$.The equation of the tangent line at $(x_1, y_1) = (0, 1)$ with slope $m = 2$ is given by $y - y_1 = m(x - x_1)$.Substituting the values: $y - 1 = 2(x - 0) \Rightarrow y = 2x + 1$.To find where the tangent meets the $x-$axis,set $y = 0$:$0 = 2x + 1 \Rightarrow 2x = -1 \Rightarrow x = -1/2$.Thus,the tangent meets the $x-$axis at the point $(-1/2, 0)$.

The tangent drawn at the point $(0, 1)$ on the curve $y = e^{2x}$ meets the $x-$axis at the point:

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