At what point does the tangent to the curve y | vedclass

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

Vedclass · Accepted Answer

$(-1/2, 0)$

Vedclass · Answer

(A) Given the curve $y = e^{2x}$.First,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 $(0, 1)$ is given by $y - y_1 = m(x - x_1)$.Substituting the values: $y - 1 = 2(x - 0) \implies y - 1 = 2x \implies 2x - y + 1 = 0$.To find where the tangent meets the $x$-axis,set $y = 0$ in the equation of the tangent line.$2x - 0 + 1 = 0 \implies 2x = -1 \implies x = -1/2$.Thus,the tangent meets the $x$-axis at the point $(-1/2, 0)$.

At what point does the tangent to the curve $y = e^{2x}$ at the point $(0, 1)$ meet the $x$-axis?

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