The slope of the normal to the curve ` y=x ^(2) +2e^(x) + 2 ` at `(0,4)` is

Options

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

Answer

To find the slope of the normal to the curve \( y = x^2 + 2e^x + 2 \) at the point \( (0, 4) \), we will follow these steps:

### Step 1: Differentiate the curve to find the slope of the tangent
The slope of the curve at any point is given by the derivative \( \frac{dy}{dx} \).

Given:
\[ y = x^2 + 2e^x + 2 \]

Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(x^2) + \frac{d}{dx}(2e^x) + \frac{d}{dx}(2) \]
Calculating each term:
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( 2e^x \) is \( 2e^x \).
- The derivative of a constant (2) is \( 0 \).

So, we have:
\[ \frac{dy}{dx} = 2x + 2e^x \]

### Step 2: Evaluate the derivative at the point \( (0, 4) \)
Now, we need to find the slope of the tangent at the point \( (0, 4) \):
\[ \frac{dy}{dx} \bigg|_{x=0} = 2(0) + 2e^0 \]
Calculating this:
\[ = 0 + 2 \cdot 1 = 2 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. If the slope of the tangent is \( m_t \), then the slope of the normal \( m_n \) is given by:
\[ m_n = -\frac{1}{m_t} \]

Substituting \( m_t = 2 \):
\[ m_n = -\frac{1}{2} \]

### Conclusion
Thus, the slope of the normal to the curve at the point \( (0, 4) \) is:
\[ \text{Slope of normal} = -\frac{1}{2} \]

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