Question

Which line(s) intersect the parabola y = x2 − 3x + 4 at two points? Select all that apply.
A. y = –3x + 2
B. y = –3x + 3
C. y = –3x + 5
D. y = –3x + 6

Answer 1

To determine which lines intersect the parabola \( y = x^2 - 3x + 4 \) at two points, we need to find the lines that satisfy the condition where the equations \( x^2 - 3x + 4 = mx + b \) (where \( m \) and \( b \) is derived from the line’s equation) have two solutions.

Rearranging \( x^2 - 3x + 4 = mx + b \) gives us:

\[
x^2 - (3 + m)x + (4 - b) = 0
\]

For this quadratic equation in \( x \) to have two intersections (real and distinct solutions), the discriminant must be positive:

\[
D = (3 + m)^2 - 4 \cdot 1 \cdot (4 - b) > 0
\]

Calculating for each option:

### A. Line: \( y = -3x + 2 \)
- Here, \( m = -3 \) and \( b = 2 \).

\[
D = (3 - 3)^2 - 4(4 - 2) = 0 - 8 = -8 \quad \text{(no intersections)}
\]

### B. Line: \( y = -3x + 3 \)
- Here, \( m = -3 \) and \( b = 3 \).

\[
D = (3 - 3)^2 - 4(4 - 3) = 0 - 4 = -4 \quad \text{(no intersections)}
\]

### C. Line: \( y = -3x + 5 \)
- Here, \( m = -3 \) and \( b = 5 \).

\[
D = (3 - 3)^2 - 4(4 - 5) = 0 + 4 = 4 \quad \text{(two intersections)}
\]

### D. Line: \( y = -3x + 6 \)
- Here, \( m = -3 \) and \( b = 6 \).

\[
D = (3 - 3)^2 - 4(4 - 6) = 0 + 8 = 8 \quad \text{(two intersections)}
\]

### Conclusion
The lines that intersect the parabola at two points are:

- **C.** \( y = -3x + 5 \)
- **D.** \( y = -3x + 6 \)

Thus, the answer is **C** and **D**.