Question

solve the system of equations by graphing. first graph the equations, and then identify the solutions.
$y = -(x - 2)^2 - 4$
$y = 4$
to graph a linear equation, click to plot points on the graph. to graph a quadratic equation, click to plot the vertex first, then plot another point on the graph.

Explanation:

Step1: Analyze the quadratic equation

The quadratic equation is \( y = -(x - 2)^2 - 4 \). The vertex form of a quadratic is \( y = a(x - h)^2 + k \), where \((h,k)\) is the vertex. Here, \(h = 2\), \(k=-4\), so the vertex is \((2, -4)\). Let's find another point. When \(x = 0\), \(y = -(0 - 2)^2 - 4 = -4 - 4=-8\), so the point \((0, -8)\) is on the parabola.

Step2: Analyze the linear equation

The linear equation is \( y = 4 \), which is a horizontal line passing through \(y = 4\) for all \(x\).

Step3: Find the intersection points

To find the solutions, we set the two equations equal: \( 4=-(x - 2)^2 - 4 \)
Add 4 to both sides: \( 8 = -(x - 2)^2 \)
Multiply both sides by -1: \( - 8=(x - 2)^2 \)
But a square of a real number can't be negative, so there are no real solutions. Wait, maybe I made a mistake. Wait, let's re - solve \(4=-(x - 2)^2-4\)
\((x - 2)^2=-8\), since the left - hand side is a square (non - negative) and the right - hand side is negative, there are no real solutions. But if we consider the graphs: the parabola \(y =-(x - 2)^2-4\) opens downward (because \(a=-1<0\)) with vertex at \((2,-4)\), and the line \(y = 4\) is above the vertex. Since the parabola opens downward and its vertex is at \(y=-4\), it will never reach \(y = 4\), so there are no real intersection points.

Answer:

There are no real solutions for the system of equations \(y=-(x - 2)^2-4\) and \(y = 4\).