Question

Consider the following system of equations:

-x + y = -2

y = -(x - 1)^2 + 5


Is 4,2 a solution to the system of equation? Explain
4,2 is (Dropdown 1: a solution, not a solution) to the system of equations because (Dropdown 2: 4,2 is a solution to only the first equation, 4,2 is a solution to both equations, 4,2 is a solution to only the second equation.


Write a graph for the system of equations above

Using your graph, state the solutions to the system of equations (give the smaller point FIRST)
__ and __

Could a system of two equations that includes a linear equation and a quadratic equation ever have no solutions? If so, how could you tell from the graph? If not, explain why. (Dropdown 1: yes, no) because (Dropdown 2: They would never not intersect, because linear and quadratic functions can never intersect.)

Answer 1

Check (4,2):

- First equation: -x + y = -2 → -4 + 2 = -2 ✓
- Second equation: y = -(x-1)^2 + 5 → -(4-1)^2 + 5 = -9 + 5 = -4 ≠ 2

So (4,2) is not a solution to the system because 4,2 is a solution to only the first equation.

Graph description:
- Line: y = x - 2
- Parabola: y = -(x-1)^2 + 5 (vertex at (1,5), opens downward)

Find intersection points algebraically:
x - 2 = -(x-1)^2 + 5
⇒ x - 2 = -x^2 + 2x + 4
⇒ 0 = -x^2 + x + 6
⇒ x^2 - x - 6 = 0
⇒ (x - 3)(x + 2) = 0 ⇒ x = 3 or x = -2

Corresponding y: y = x - 2 gives points (3,1) and (-2,-4).

Solutions (smaller point first): (-2, -4) and (3, 1).

Could there be no solutions? Yes — because the line and the parabola might not intersect. On the graph you would see no intersection points (equivalently, solving for x gives a quadratic with no real roots — discriminant < 0).