The tangent to the curve ` y = e^(2x) ` at the point (0,1) meets 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: Find the slope of the tangent line
The slope of the tangent line to the curve at a point is given by the derivative of the function at that point.

The function is:
\[ y = e^{2x} \]

To find the derivative, we use the chain rule:
\[ \frac{dy}{dx} = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x} \]

### Step 2: Evaluate the derivative at the point \( (0, 1) \)
Now, we substitute \( x = 0 \) into the derivative to find the slope at that point:
\[ \frac{dy}{dx} \bigg|_{x=0} = 2e^{2 \cdot 0} = 2e^{0} = 2 \cdot 1 = 2 \]

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

### Step 4: Find where the tangent line meets the x-axis
The x-axis is defined by \( y = 0 \). To find the x-coordinate where the tangent line meets the x-axis, we set \( y = 0 \) in the tangent line equation:
\[ 0 = 2x + 1 \]

### Step 5: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = -1 \]
\[ x = -\frac{1}{2} \]

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

### Final Answer
The tangent meets the x-axis at the point \( \left(-\frac{1}{2}, 0\right) \).

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