1. We are asked to find solutions to inequalities by graphing three parabolas and analyzing their vertices and tables of values. 2. For the first inequality: $y > (x-5)^2 + 2$ - The vertex is at $(5, 2)$. - The parabola opens upward since the coefficient of the squared term $(x-5)^2$ is positive. - Values near the vertex: For $x=4$, $y=(4-5)^2+2=1+2=3$; for $x=6$, $y=(6-5)^2+2=1+2=3$. - So, points near vertex are $(4,3)$ and $(6,3)$. - The solution region is above the parabola since $y >$ the parabola. 3. For the second inequality: $y \leq 2(x+1)^2 - 4$ - The vertex is at $(-1, -4)$. - The parabola opens upward and is vertically stretched by factor 2. - Calculate nearby points: for $x=0$, $y=2(0+1)^2 - 4 = 2(1) - 4 = -2$; for $x=-2$, $y=2(-2+1)^2 - 4 = 2(1)^2 - 4 = -2$. - Points near vertex: $(0,-2)$ and $(-2,-2)$. - The solution region is on or below the parabola since $y \leq$ the parabola. 4. For the third inequality: $y \geq x^2 + 8x + 25$ - Complete the square: $x^2 + 8x + 25 = (x^2 + 8x + 16) + 9 = (x+4)^2 + 9$ - Vertex is at $(-4, 9)$. - Parabola opens upward. - Calculate values: for $x=-3$, $y=(-3+4)^2+9=1+9=10$; for $x=-5$, $y=(-5+4)^2+9=1+9=10$. - Nearby points: $(-3,10)$ and $(-5,10)$. - Solution region is on or above the parabola since $y \geq$ the parabola. 5. Summary: - Inequality 1: vertex $(5,2)$, solutions where $y > (x-5)^2 + 2$, i.e., points above parabola. - Inequality 2: vertex $(-1,-4)$, solutions where $y \leq 2(x+1)^2 - 4$, points on or below parabola. - Inequality 3: vertex $(-4,9)$, solutions where $y \geq (x+4)^2 + 9$, points on or above parabola. These vertices and values provide key reference points to graph and identify solution regions.