The slopes of the tangents to the curve `y=(x+1)(x-3)` at the points where it cuts the x - axis, are `m_(1) and m_(2)`, then the value of `m_(1)+m_(2)` is equal to

Options

A. 8
B. `-2`
C. 2
D. 0

Answer

To solve the problem, we need to find the slopes of the tangents to the curve \( y = (x + 1)(x - 3) \) at the points where it intersects the x-axis, and then calculate the sum of these slopes.

### Step 1: Find the points where the curve intersects the x-axis
The curve intersects the x-axis where \( y = 0 \). Therefore, we set the equation to zero:

\[ 0 = (x + 1)(x - 3) \]

### Step 2: Solve for \( x \)
To find the values of \( x \), we solve the equation:

\[ (x + 1)(x - 3) = 0 \]

This gives us two solutions:

1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x - 3 = 0 \) → \( x = 3 \)

Thus, the points where the curve intersects the x-axis are \( x = -1 \) and \( x = 3 \).

### Step 3: Differentiate the curve
Next, we need to find the derivative of the function to determine the slopes of the tangents at these points. The function is:

\[ y = (x + 1)(x - 3) \]

Using the product rule, we differentiate:

\[ \frac{dy}{dx} = (x - 3) \cdot \frac{d}{dx}(x + 1) + (x + 1) \cdot \frac{d}{dx}(x - 3) \]

Calculating the derivatives:

\[ \frac{d}{dx}(x + 1) = 1 \quad \text{and} \quad \frac{d}{dx}(x - 3) = 1 \]

Substituting back, we get:

\[ \frac{dy}{dx} = (x - 3)(1) + (x + 1)(1) = (x - 3) + (x + 1) = 2x - 2 \]

### Step 4: Calculate the slopes at the intersection points
Now we will find the slopes \( m_1 \) and \( m_2 \) at \( x = -1 \) and \( x = 3 \) respectively.

1. For \( x = -1 \):

\[ m_1 = \frac{dy}{dx} \bigg|_{x = -1} = 2(-1) - 2 = -2 - 2 = -4 \]

2. For \( x = 3 \):

\[ m_2 = \frac{dy}{dx} \bigg|_{x = 3} = 2(3) - 2 = 6 - 2 = 4 \]

### Step 5: Calculate \( m_1 + m_2 \)
Now we can find the sum of the slopes:

\[ m_1 + m_2 = -4 + 4 = 0 \]

### Final Answer
The value of \( m_1 + m_2 \) is \( \boxed{0} \).