The equation of tangent to the curve `y=x^(2)+4x+1` at (-1, -2) is

Options

A. `2x-y=0`
B. `2x+y-5=0`
C. `2x-y-1=0`
D. `x+y-1=0`.

Answer

To find the equation of the tangent to the curve \( y = x^2 + 4x + 1 \) at the point \((-1, -2)\), we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative of the function \( y \) with respect to \( x \) to determine the slope of the tangent line.

\[ y = x^2 + 4x + 1 \]

Differentiating \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 2x + 4 \]

### Step 2: Evaluate the derivative at the given point
Now, we will evaluate the derivative at the point \( x = -1 \) to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=-1} = 2(-1) + 4 = -2 + 4 = 2 \]

### Step 3: Use the point-slope form of the line
We have the slope \( m = 2 \) and the point \((-1, -2)\). We can use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting the values:

\[ y - (-2) = 2(x - (-1)) \]

This simplifies to:

\[ y + 2 = 2(x + 1) \]

### Step 4: Simplify the equation
Now, we will simplify the equation:

\[ y + 2 = 2x + 2 \]

Subtracting 2 from both sides gives:

\[ y = 2x + 2 - 2 \]

Thus, we have:

\[ y = 2x \]

### Step 5: Rearranging to standard form
To express the equation in standard form, we can rearrange it:

\[ 2x - y = 0 \]

### Final Answer
The equation of the tangent to the curve at the point \((-1, -2)\) is:

\[ 2x - y = 0 \]

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