The equation of a curve is given as `y= x^(2) + 2-3x`. The curve intersects the x-axis at

Options

A. `(1, 0)`
B. `(2, 0) `
C. Both (1) and (2)
D. No where

Explanation

To find the points where the curve intersects the x-axis, we need to set \( y = 0 \) in the given equation \( y = x^2 + 2 - 3x \).

### Step-by-Step Solution:

1. **Set the equation to zero**:
\[ 0 = x^2 + 2 - 3x \]

2. **Rearrange the equation**:
\[ x^2 - 3x + 2 = 0 \]

3. **Factor the quadratic equation**:
We need to factor \( x^2 - 3x + 2 \). To do this, we look for two numbers that multiply to \( 2 \) (the constant term) and add up to \( -3 \) (the coefficient of \( x \)).
- The numbers are \( -1 \) and \( -2 \).
- Thus, we can factor the equation as:
\[ (x - 1)(x - 2) = 0 \]

4. **Set each factor to zero**:
\[ x - 1 = 0 \quad \text{or} \quad x - 2 = 0 \]

5. **Solve for \( x \)**:
- From \( x - 1 = 0 \), we get \( x = 1 \).
- From \( x - 2 = 0 \), we get \( x = 2 \).

6. **Conclusion**:
The curve intersects the x-axis at the points \( x = 1 \) and \( x = 2 \).

### Final Answer:
The curve intersects the x-axis at the points \( (1, 0) \) and \( (2, 0) \).

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