Question

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Solve the system of equations by graphing. First graph the equations, and then identify the solutions.
y
=
–
(x+3)2+1
y
=
1
2
x+5
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.
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What are the solutions to the system of equations? Select all that apply.

Answer 1

To solve the system of equations by graphing, we will first examine the given equations:

1. **Quadratic equation:** \( y = - (x + 3)^2 + 1 \)
2. **Linear equation:** \( y = \frac{1}{2} x + 5 \)

### Step 1: Graphing the Quadratic Equation

1. **Find the vertex:** The vertex form of the quadratic function is \( y = a(x-h)^2 + k \). Here, \( (h, k) \) is the vertex.
- The vertex is at \( (-3, 1) \).

2. **Plot the vertex**: So we plot the point \( (-3, 1) \).

3. **Find another point:** Choose a value for \( x \) to find a corresponding \( y \):
- For \( x = -2 \):
\[
y = -(-2 + 3)^2 + 1 = -1 + 1 = 0 \quad \text{(Point: } (-2, 0) \text{)}
\]
- For \( x = -4 \):
\[
y = -(-4 + 3)^2 + 1 = -1 + 1 = 0 \quad \text{(Point: } (-4, 0) \text{)}
\]

4. **Draw the parabola**: Plot the points \( (-3, 1) \), \( (-2, 0) \), and \( (-4, 0) \) and draw the parabola opening downwards.

### Step 2: Graphing the Linear Equation

1. **Find the y-intercept**: Set \( x = 0 \):
\[
y = \frac{1}{2}(0) + 5 = 5 \quad \text{(Point: } (0, 5) \text{)}
\]

2. **Find another point**: For \( x = -2 \):
\[
y = \frac{1}{2}(-2) + 5 = -1 + 5 = 4 \quad \text{(Point: } (-2, 4) \text{)}
\]

3. **Draw the line**: Plot the points \( (0, 5) \) and \( (-2, 4) \) and draw the line.

### Step 3: Find the Intersection Points

To solve the system, we need to find the points where the graphs of the two equations intersect.

- **Graphically**, check where the parabola intersects the line. It appears (after plotting) that there are two points of intersection:

1. One point around \( (-3, 1) \)
2. Another point around \( (-2, 0) \)

### Conclusion

The solutions to the system of equations, identified by the intersection points on the graph, are:
- Approximately \( (-3, 1) \)
- Approximately \( (-2, 4) \)

So, the solutions to the system of equations are:
1. \( (-3, 1) \)
2. \( (-2, 4) \)

Please check with actual plotting to confirm exact intersection points.