You list only one line: y = -3x - 3, so that 's all we can check.
If the parabola is represented by y – x + 2 = x^2, then replace y here by
y = -3x - 3:
-3x - 3 – x + 2 = x^2. Then determine whether or not this new equation has a real solution:
Combining like terms, -3x - x -3 + 2 = x^2, or
-4x -1 = x^2, or x^2 + 4x + 1 = 0.
Does this have real solutions? Find the determinant, b^2 - 4ac:
determinant = (4)^2 - 4(1)(1) = 16 - 4 = 12
Thus, this quadratic equation will have two real, unequal roots. At least one of these roots should be the x-coordinate of the point of intersection(s) of the given parabola and line.
You'll need to check out your other answer choices using the same or a similar method.
