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 equation of the curve. The equation given is:

\[ y = x^2 + 2 - 3x \]

### Step 1: Set the equation to zero
To find the x-intercepts, we set \( y \) to zero:

\[ 0 = x^2 + 2 - 3x \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ x^2 - 3x + 2 = 0 \]

### Step 3: Factor the quadratic equation
Next, we will factor the quadratic equation. We need two numbers that multiply to \( 2 \) (the constant term) and add up to \( -3 \) (the coefficient of \( x \)). The numbers \( -1 \) and \( -2 \) satisfy these conditions:

\[ (x - 1)(x - 2) = 0 \]

### Step 4: Solve for x
Now, we can set each factor equal to zero:

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

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

### Final Answer
The points of intersection are \( (1, 0) \) and \( (2, 0) \).

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