Solve System of Inequalities $ y \leq (x + 1)^2 - 4 $ and $ y 2x + 1 $ (Algebra 2)

Question

Sketch the solution to the system of inequalities $ y \leq (x + 1)^2 - 4 $ and $ y > 2x + 1 $, then identify one solution to the system.

Accepted Answer

One solution is $(-3, -2)$ (other valid points exist).

Answer

1. Analyze $ y \leq (x + 1)^2 - 4 $: The quadratic $ (x + 1)^2 - 4 $ is a parabola opening upward with vertex at $(-1, -4)$. The inequality $ \leq $ means the parabola is solid, and we shade below it. 2. Analyze $ y > 2x + 1 $: This is a linear line with slope $ 2 $ and $ y $-intercept $ 1 $. The inequality $ > $ means the line is dashed, and we shade above it. 3. Identify the overlapping region (solution): Solve $ (x + 1)^2 - 4 = 2x + 1 $ to find intersection points: $ x^2 - 4 = 0 $ → $ x = 2 $ or $ x = -2 $ (points $(2, 5)$ and $(-2, -3)$). Test the point $(-3, -2)$:    - For $ y \leq (-3 + 1)^2 - 4 $: $ -2 \leq 4 - 4 $ → $ -2 \leq 0 $ (true).    - For $ y > 2(-3) + 1 $: $ -2 > -6 + 1 $ → $ -2 > -5 $ (true). Thus, $(-3, -2)$ is a solution.