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 \( y = 2 - x \):
- For \( x = -4 \):
\[ y = 2 - (-4) = 2 + 4 = 6 \]
So one intersection point is \( (-4, 6) \).
- 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:
\[ (-4, 6) \quad \text{and} \quad (1, 1) \]
