Graph Inequalities A60E46

1. **Problem:** Graph the inequality $y \geq x^2 - 2$. **Step 1:** Identify the boundary curve: $y = x^2 - 2$ is a parabola opening upwards with vertex at $(0,-2)$. **Step 2:** Since the inequality is $\geq$, shade the region above or on the parabola. **Step 3:** The parabola is solid because points on it satisfy the inequality. 2. **Problem:** Graph the inequality $y < x^2 + 2x - 3$. **Step 1:** Boundary curve: $y = x^2 + 2x - 3$ is a parabola opening upwards. **Step 2:** Rewrite in vertex form: $$y = (x+1)^2 - 4$$ vertex at $(-1,-4)$. **Step 3:** Since inequality is $<$, shade below the parabola. **Step 4:** Use a dashed curve because points on the parabola are not included. 3. **Problem:** Graph the inequality $x^2 + y \leq 16$. **Step 1:** Rewrite as $y \leq 16 - x^2$. **Step 2:** Boundary curve: $y = 16 - x^2$ is a downward-opening parabola with vertex at $(0,16)$. **Step 3:** Shade the region below or on the parabola. **Step 4:** Solid curve because $\leq$ includes boundary. 4. **Problem:** Graph the inequality $y > -x^2 + 6$. **Step 1:** Boundary curve: $y = -x^2 + 6$ is a downward-opening parabola with vertex at $(0,6)$. **Step 2:** Since inequality is $>$, shade above the parabola. **Step 3:** Use a dashed curve because points on the parabola are not included. 5. **Problem:** Graph the inequality $y \leq x^2 - 4x + 1$. **Step 1:** Rewrite in vertex form: $$y = (x^2 - 4x + 4) - 3 = (x-2)^2 - 3$$ vertex at $(2,-3)$. **Step 2:** Since inequality is $\leq$, shade below or on the parabola. **Step 3:** Use a solid curve because boundary points are included. **Summary:** - For $\geq$ or $\leq$, use solid boundary curves and shade above or below accordingly. - For $>$ or $<$, use dashed boundary curves and shade above or below accordingly. - Convert quadratic expressions to vertex form to identify vertex and direction. This completes the solutions for all five graphing inequalities.