Question

select all the solutions to the system of equations.
(-4, 0) (-2, 8)
(-1, 9) (0, 8)
(2, 0)

Explanation:

Step1: Recall solution definition

Solutions to a system of equations (represented by graphs) are the intersection points of the graphs. So we need to find the points where the two curves (the line and the parabola) intersect.

Step2: Analyze each point

- For \((-4, 0)\): Check if both graphs pass through this point. From the graph, the parabola and the line intersect here? Wait, no—wait, the parabola and the line: let's look at the graph. Wait, actually, the solution of a system of equations (graphically) is where the two graphs intersect. So let's check each point:

1. \((-4, 0)\): Does the line and the parabola both pass through \((-4, 0)\)? Looking at the graph, the parabola and the line intersect at \((-4, 0)\)? Wait, no, wait the line and the parabola: let's see the intersection points. Wait, the two graphs (the line and the parabola) intersect at three points? Wait, no, the line is a straight line, the parabola is a quadratic. Wait, actually, the solution to the system is the set of \((x, y)\) that satisfy both equations, i.e., the intersection points of the two graphs.

Looking at the graph:

- \((0, 8)\): Both the line and the parabola pass through \((0, 8)\) (since they intersect at the y-axis there).
- \((-4, 0)\): Wait, does the line pass through \((-4, 0)\)? Let's check the line: it goes through \((-4, 0)\) and \((0, 8)\), so slope is \(\frac{8 - 0}{0 - (-4)} = 2\), equation \(y = 2x + 8\). The parabola: let's see, it passes through \((-4, 0)\), \((0, 8)\), \((2, 0)\). So the parabola is a quadratic with roots at \(x = -4\) and \(x = 2\), so equation \(y = a(x + 4)(x - 2)\). Plugging in \((0, 8)\): \(8 = a(4)(-2) \implies 8 = -8a \implies a = -1\), so \(y = - (x + 4)(x - 2) = -x^2 - 2x + 8\). Now check the line \(y = 2x + 8\) and parabola \(y = -x^2 - 2x + 8\). Set equal: \(2x + 8 = -x^2 - 2x + 8 \implies x^2 + 4x = 0 \implies x(x + 4) = 0 \implies x = 0\) or \(x = -4\). Wait, so solutions are \(x = 0\) (y=8) and \(x = -4\) (y=0), and wait, no—wait, that's a quadratic equation, so two solutions? But the graph shows three points? Wait, no, maybe I made a mistake. Wait, the line is \(y = 2x + 8\), the parabola is \(y = -x^2 - 2x + 8\). Solving \(2x + 8 = -x^2 - 2x + 8\) gives \(x^2 + 4x = 0 \implies x(x + 4) = 0 \implies x = 0\) or \(x = -4\). So the solutions are \((0, 8)\) and \((-4, 0)\)? Wait, but the graph also shows the parabola's vertex at \((-1, 9)\), but does the line pass through \((-1, 9)\)? Let's check the line: \(y = 2(-1) + 8 = 6\), which is not 9. So \((-1, 9)\) is only on the parabola, not the line. \((-2, 8)\): line at \(x = -2\): \(y = 2(-2) + 8 = 4\), not 8. So \((-2, 8)\) is not on the line. \((2, 0)\): line at \(x = 2\): \(y = 2(2) + 8 = 12\), not 0. So the correct solutions are \((-4, 0)\) and \((0, 8)\), and wait, no—wait, when we solved the equations, we got \(x = 0\) and \(x = -4\), so \((0, 8)\) and \((-4, 0)\). Wait, but the initial check: the user's options had \((-4, 0)\), \((-2, 8)\), \((-1, 9)\), \((0, 8)\), \((2, 0)\). But according to the equations, the solutions are \((-4, 0)\) and \((0, 8)\), and wait, did I make a mistake? Wait, let's re-express:

Wait, the line is \(y = 2x + 8\) (since it goes from \((-4, 0)\) to \((0, 8)\)). The parabola: let's check \((-1, 9)\): plug into parabola equation \(y = -(-1)^2 - 2(-1) + 8 = -1 + 2 + 8 = 9\), so \((-1, 9)\) is on the parabola, but is it on the line? Line at \(x = -1\): \(y = 2(-1) + 8 = 6 eq 9\), so not on the line. \((-2, 8)\): line at \(x = -2\): \(y = 2(-2) + 8 = 4 eq 8\), so not on the line. \((2, 0)\): line at \(x = 2\): \(y = 2(2) + 8 = 12 eq 0\), so not on the line. So the correct solutions are \((-4, 0)\) and \((0, 8)\), and wait, but the user's options: let's check the graph again. Wait, maybe I misread the graphs. Wait, the two graphs (the line and the parabola) intersect at three points? No, a line and a parabola can intersect at 0, 1, or 2 points. Wait, the line \(y = 2x + 8\) and parabola \(y = -x^2 - 2x + 8\): solving gives \(2x + 8 = -x^2 - 2x + 8 \implies x^2 + 4x = 0 \implies x(x + 4) = 0\), so two solutions: \(x = 0\) (y=8) and \(x = -4\) (y=0). So the correct solutions are \((-4, 0)\) and \((0, 8)\), and wait, but the user's options have \((0, 8)\) as a choice (the circle is around it, maybe initially unchecked). Wait, the original problem's options: the user has some checked, but we need to select all correct.

So:

- \((-4, 0)\): correct (intersection of line and parabola)
- \((-2, 8)\): not correct (line at x=-2 is y=4, not 8)
- \((-1, 9)\): not correct (line at x=-1 is y=6, not 9)
- \((0, 8)\): correct (intersection of line and parabola)
- \((2, 0)\): not correct (line at x=2 is y=12, not 0)

Wait, but the user's initial check marks: maybe a mistake. Let's re-express:

The solution to the system of equations (graphically) is the set of points where the two graphs (the line and the parabola) intersect. From the graph, these points are:

1. \((-4, 0)\): both graphs pass through here.
2. \((0, 8)\): both graphs pass through here.

Wait, but the parabola also passes through \((2, 0)\), but the line does not, so \((2, 0)\) is not a solution. \((-2, 8)\): line does not pass through here (line at x=-2 is y=4), parabola at x=-2: \(y = -(-2)^2 - 2(-2) + 8 = -4 + 4 + 8 = 8\), so parabola passes through \((-2, 8)\), but line does not, so \((-2, 8)\) is not a solution. \((-1, 9)\): parabola passes through here (vertex), line does not (line at x=-1 is y=6), so not a solution.

So the correct solutions are \((-4, 0)\) and \((0, 8)\).

Answer:

\((-4, 0)\), \((0, 8)\)