Answer :
To solve the system of equations defined by the line [tex]\(2x + y = -4\)[/tex] and the parabola [tex]\(y = (x + 1)^2 - 2\)[/tex], we follow these steps:
### Step 1: Express Both Equations Clearly
The line equation is:
[tex]\[ 2x + y = -4 \][/tex]
The parabola equation is:
[tex]\[ y = (x + 1)^2 - 2 \][/tex]
### Step 2: Substitute in One Equation
Solve the line equation for [tex]\(y\)[/tex]:
From [tex]\(2x + y = -4\)[/tex]:
[tex]\[ y = -4 - 2x \][/tex]
### Step 3: Substitute [tex]\(y\)[/tex] into the Second Equation
Substitute [tex]\(y\)[/tex] from the line equation into the parabola equation:
[tex]\[ -4 - 2x = (x + 1)^2 - 2 \][/tex]
### Step 4: Expand and Simplify
Expand and simplify the quadratic equation:
[tex]\[ -4 - 2x = x^2 + 2x + 1 - 2 \][/tex]
[tex]\[ -4 - 2x = x^2 + 2x - 1 \][/tex]
Combine all terms into one side to form a standard quadratic equation:
[tex]\[ x^2 + 2x - 1 + 2x + 4 = 0 \][/tex]
[tex]\[ x^2 + 4x + 3 = 0 \][/tex]
### Step 5: Solve the Quadratic Equation
Solve for [tex]\(x\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 3\)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2}{2} \][/tex]
So the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = \frac{-4 + 2}{2} = -1 \][/tex]
[tex]\[ x = \frac{-4 - 2}{2} = -3 \][/tex]
### Step 6: Find Corresponding [tex]\(y\)[/tex] Values
Substitute [tex]\(x = -1\)[/tex] and [tex]\(x = -3\)[/tex] back into [tex]\( y = -4 - 2x \)[/tex] to find corresponding [tex]\(y\)[/tex] values:
For [tex]\(x = -1\)[/tex]:
[tex]\[ y = -4 - 2(-1) \][/tex]
[tex]\[ y = -4 + 2 \][/tex]
[tex]\[ y = -2 \][/tex]
For [tex]\(x = -3\)[/tex]:
[tex]\[ y = -4 - 2(-3) \][/tex]
[tex]\[ y = -4 + 6 \][/tex]
[tex]\[ y = 2 \][/tex]
### Results:
The solutions to the system, which are the points where the line intersects the parabola, are:
[tex]\[ (-3, 2) \][/tex]
[tex]\[ (-1, -2) \][/tex]
Thus, the intersection points are:
[tex]\[ \boxed{(-3, 2) \text{ and } (-1, -2)} \][/tex]
### Step 1: Express Both Equations Clearly
The line equation is:
[tex]\[ 2x + y = -4 \][/tex]
The parabola equation is:
[tex]\[ y = (x + 1)^2 - 2 \][/tex]
### Step 2: Substitute in One Equation
Solve the line equation for [tex]\(y\)[/tex]:
From [tex]\(2x + y = -4\)[/tex]:
[tex]\[ y = -4 - 2x \][/tex]
### Step 3: Substitute [tex]\(y\)[/tex] into the Second Equation
Substitute [tex]\(y\)[/tex] from the line equation into the parabola equation:
[tex]\[ -4 - 2x = (x + 1)^2 - 2 \][/tex]
### Step 4: Expand and Simplify
Expand and simplify the quadratic equation:
[tex]\[ -4 - 2x = x^2 + 2x + 1 - 2 \][/tex]
[tex]\[ -4 - 2x = x^2 + 2x - 1 \][/tex]
Combine all terms into one side to form a standard quadratic equation:
[tex]\[ x^2 + 2x - 1 + 2x + 4 = 0 \][/tex]
[tex]\[ x^2 + 4x + 3 = 0 \][/tex]
### Step 5: Solve the Quadratic Equation
Solve for [tex]\(x\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 3\)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2}{2} \][/tex]
So the solutions for [tex]\(x\)[/tex] are:
[tex]\[ x = \frac{-4 + 2}{2} = -1 \][/tex]
[tex]\[ x = \frac{-4 - 2}{2} = -3 \][/tex]
### Step 6: Find Corresponding [tex]\(y\)[/tex] Values
Substitute [tex]\(x = -1\)[/tex] and [tex]\(x = -3\)[/tex] back into [tex]\( y = -4 - 2x \)[/tex] to find corresponding [tex]\(y\)[/tex] values:
For [tex]\(x = -1\)[/tex]:
[tex]\[ y = -4 - 2(-1) \][/tex]
[tex]\[ y = -4 + 2 \][/tex]
[tex]\[ y = -2 \][/tex]
For [tex]\(x = -3\)[/tex]:
[tex]\[ y = -4 - 2(-3) \][/tex]
[tex]\[ y = -4 + 6 \][/tex]
[tex]\[ y = 2 \][/tex]
### Results:
The solutions to the system, which are the points where the line intersects the parabola, are:
[tex]\[ (-3, 2) \][/tex]
[tex]\[ (-1, -2) \][/tex]
Thus, the intersection points are:
[tex]\[ \boxed{(-3, 2) \text{ and } (-1, -2)} \][/tex]
