Which line(s) intersect the parabola y = x^2 − 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

C and D

Step-by-step explanation:

Equating the line A and the parabola, we get

-3x + 2 = x² - 3x + 4

0 = x² - 3x + 4 +3x - 2

0 = x² + 2

-2 = x²

which has no real solutions. Then, the line A and the parabola don't intersect each other.

Equating the line B and the parabola, we get

-3x + 3 = x² - 3x + 4

0 = x² - 3x + 4 + 3x - 3

0 = x² + 1

-1 = x²

which has no real solutions. Then, the line B and the parabola don't intersect each other.

Equating the line C and the parabola, we get

-3x + 5 = x² - 3x + 4

0 = x² - 3x + 4  + 3x - 5

0 = x² - 1

1 = x²

√1 = x

which has 2 solutions, x = 1 and x = -1. Then, the line C and the parabola intersect each other.

Equating the line D and the parabola, we get

-3x + 6 = x² - 3x + 4

0 = x² - 3x + 4  + 3x - 6

0 = x² - 2

2 = x²

√2 = x

which has 2 solutions, x ≈ 1.41 and x ≈ -1.41. Then, the line D and the parabola intersect each other.

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents