The tangent to the curve y = e^{2x} at the po | vedclass

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(C) Given the curve $y = e^{2x}$.First,we find the derivative $\frac{dy}{dx} = 2e^{2x}$.At the point $(0, 1)$,the slope of the tangent is $m = \left(\

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$(-\frac{1}{2}, 0)$

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

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

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