To find the intersection points of the given equations:

1. Equations:

   • $y = 4 - x^2$ (Equation 1)
   • $y = x + 2$ (Equation 2)

2. Find intersection points by equating the two equations:

   Set $4 - x^2 = x + 2$: $4 - x^2 = x + 2$

3. Simplify the equation: Bring all terms to one side: $-x^2 - x + 4 - 2 = 0$ $-x^2 - x + 2 = 0$ Multiply through by $-1$ to simplify: $x^2 + x - 2 = 0$

4. Solve the quadratic equation: Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 1$, and $c = -2$: $x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)}$ $x = \frac{-1 \pm \sqrt{1 + 8}}{2}$ $x = \frac{-1 \pm \sqrt{9}}{2}$ $x = \frac{-1 \pm 3}{2}$

   Solve for the two roots: $x = \frac{-1 + 3}{2} = 1 \quad \text{and} \quad x = \frac{-1 - 3}{2} = -2$

5. Find the corresponding $y$-values: Substitute $x = 1$ and $x = -2$ into $y = x + 2$ (Equation 2):

   • For $x = 1$: $y = 1 + 2 = 3$ So, one intersection point is $(1, 3)$.
   • For $x = -2$: $y = -2 + 2 = 0$ So, the other intersection point is $(-2, 0)$.

6. Intersection points: $\boxed{(1, 3) \quad \text{and} \quad (-2, 0)}$

Would you like a graph of the intersections or further explanation? 😊

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5 Related Questions:

1. How do you graphically represent these intersections on a coordinate plane?
2. What happens if you change $y = 4 - x^2$ to $y = 4 - 2x^2$?
3. How can you confirm the solutions algebraically using substitution?
4. Are these intersections also solutions to other equations derived from these?
5. What is the significance of solving quadratic equations in real-world applications?

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Tip:

Always double-check your solutions by substituting the $x$-values back into both original equations to ensure they satisfy them!