Question

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

• A. (0,1)
• B. (2, 0)
• C. (-12, 0)
• D. (-2, 0)

Answer 1

The tangent to the curve $y=e^{2x}$ at the point (0, 1) meets the x-axis at $\left(-\frac{1}{2}, 0\right)$. None of the provided options (A, B, C, D) match this result.

Answer 2

Explanation:

1. Find the derivative of the curve: Given the curve $y = e^{2x}$. Differentiate with respect to $x$:

   $$\frac{dy}{dx} = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x}$$

2. Calculate the slope of the tangent at the given point (0, 1): Substitute $x=0$ into the derivative to find the slope $m$:

   $$m = \frac{dy}{dx} \Big|_{(0,1)} = 2e^{2(0)} = 2e^0 = 2(1) = 2$$

3. Determine the equation of the tangent line: Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, with point $(x_1, y_1) = (0, 1)$ and slope $m=2$:

   $$y - 1 = 2(x - 0)$$ $$y - 1 = 2x$$ $$y = 2x + 1$$

4. Find the x-intercept of the tangent line: The tangent line meets the x-axis when $y=0$. Substitute $y=0$ into the tangent equation:

   $$0 = 2x + 1$$ $$2x = -1$$ $$x = -\frac{1}{2}$$

   Thus, the tangent meets the x-axis at the point $\left(-\frac{1}{2}, 0\right)$.

5. Compare with given options: The calculated point is $\left(-\frac{1}{2}, 0\right)$. The given options are: A) (0,1) B) (2, 0) C) (-12, 0) D) (-2, 0) None of the provided options match the calculated correct point $\left(-\frac{1}{2}, 0\right)$.