The tangent to the curve y = e2x at the point (0, 1) meets x-axis at ______.

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The tangent to the curve y = e2x at the point (0, 1) meets x-axis at (- $\frac{1}{2}$, 0). Explanation: Equation of the curve is y = e2x Slope of the tangent "dy"/"dx" = 2e2x $\Rightarrow$ "dy"/"dx"_(0, 1) = 2 · e0 = 2 $\therefore$ Equation of tangent to the curve at (0, 1) is y –1 = 2(x – 0) $\Rightarrow$ y – 1 = 2x $\Rightarrow$ y – 2x = 1 Since the tangent meets x-axis where y = 0 $\therefore$ 0 – 2x = 1 $\Rightarrow$ x = (-1)/2 So the point is (- $\frac{1}{2}$, 0).

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