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

Options

A. (0,1)
B. `(-(1)/(2),0)`
C. (2,0)
D. (0,2)

Answer

To find the point where the tangent to the curve \( y = e^{2x} \) at the point (0,1) meets the x-axis, we can follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = e^{2x} \]
To find the slope of the tangent line at any point, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{2x}) = 2e^{2x} \]

### Step 2: Evaluate the derivative at the point (0,1)
Next, we evaluate the derivative at the point \( x = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = 2e^{2 \cdot 0} = 2e^{0} = 2 \cdot 1 = 2 \]
Thus, the slope of the tangent line at the point (0,1) is 2.

### Step 3: Write the equation of the tangent line
Using the point-slope form of the equation of a line, we can write the equation of the tangent line:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope and \( (x_1, y_1) \) is the point (0,1):
\[ y - 1 = 2(x - 0) \]
This simplifies to:
\[ y = 2x + 1 \]

### Step 4: Find where the tangent meets the x-axis
To find where this tangent line meets the x-axis, we set \( y = 0 \):
\[ 0 = 2x + 1 \]
Solving for \( x \):
\[ 2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2} \]

### Step 5: Write the coordinates of the intersection point
The point where the tangent meets the x-axis is:
\[ \left(-\frac{1}{2}, 0\right) \]

### Final Answer
Thus, the tangent to the curve \( y = e^{2x} \) at the point (0,1) meets the x-axis at:
\[ \boxed{\left(-\frac{1}{2}, 0\right)} \]