Y=x^2+2x-2 y=2-x

To find the intersection points of the equations \( y = x^2 + 2x - 2 \) and \( y = 2 - x \), we can set them equal to each other: \[ x^2 + 2x - 2 = 2 - x \] Now, rearranging the equation: \[ x^2 + 2x + x - 2 - 2 = 0 \] This simplifies to: \[ x^2 + 3x - 4 = 0 \] Next, we can factor this quadratic equation. We are looking for two numbers that multiply to \(-4\) and add to \(3\). The numbers \(4\) and \(-1\) work, so we can factor it as: \[ (x + 4)(x - 1) = 0 \] Setting each factor equal to zero gives us the possible \(x\) values: \[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Now we can find the corresponding \(y\) values for each \(x\) value using either of the original equations. We'll use \( y = 2 - x \): 1. For \( x = -4 \): \[ y = 2 - (-4) = 2 + 4 = 6 \] So one intersection point is \( (-4, 6) \). 2. For \( x = 1 \): \[ y = 2 - 1 = 1 \] So the other intersection point is \( (1, 1) \). Therefore, the intersection points of the equations \( y = x^2 + 2x - 2 \) and \( y = 2 - x \) are: \[ (-4, 6) \quad \text{and} \quad (1, 1) \]