Let's solve the system of equations by graphing.

The given equations are:

1. $y = 2(x + 3)^2 - 1$ (a quadratic equation)
2. $y = -2x - 3$ (a linear equation)

Step 1: Graph the quadratic equation $y = 2(x + 3)^2 - 1$

This is in vertex form $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex.

• The vertex is at $(-3, -1)$, since $h = -3$ and $k = -1$.
• The coefficient $a = 2$ tells us the parabola opens upward (since it's positive) and the "stretch" factor is 2, which means it will be narrower than a standard parabola.

Step 2: Graph the linear equation $y = -2x - 3$

For the linear equation, we can identify two key points:

• The y-intercept is $-3$, so the point $(0, -3)$ is on the line.
• The slope is $-2$, which means the line goes down 2 units for every 1 unit moved to the right. So, starting from the point $(0, -3)$, the next point will be $(1, -5)$.

Step 3: Graph the equations

1. For the quadratic, plot the vertex $(-3, -1)$, and then choose a few more x-values (such as $x = -4, -2, 0$) to calculate the corresponding y-values and sketch the curve.
2. For the linear equation, plot the points $(0, -3)$ and $(1, -5)$, and then draw a straight line through them.

Step 4: Find the intersection point

The solution to the system is where the quadratic and linear graphs intersect. You can find this point by solving the equations algebraically or by identifying it visually from the graph.

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Would you like me to generate the graphs to show you the solution visually?