Inequality Graphing 50Afda

1. **State the problem:** We need to graph and analyze the system of inequalities: $$y > x^2 - 25$$ $$y \geq x^2 - 4x + 4$$ $$y < x^2 - 5x + 6$$ $$y \leq -x^2 + 5x + 4$$ $$3x - 4y \geq 12$$ 2. **Features of each parabola and line:** - For each quadratic inequality, identify the vertex, axis of symmetry, and direction of opening. - For the linear inequality, rewrite in slope-intercept form. 3. **Analyze each inequality:** **Inequality 1:** $$y > x^2 - 25$$ - This is a parabola opening upwards. - Vertex at $$ (0, -25) $$. - The region is above the parabola. **Inequality 2:** $$y \geq x^2 - 4x + 4$$ - Rewrite quadratic: $$x^2 - 4x + 4 = (x-2)^2$$. - Vertex at $$ (2, 0) $$. - Region is on or above this parabola. **Inequality 3:** $$y < x^2 - 5x + 6$$ - Factor quadratic: $$x^2 - 5x + 6 = (x-2)(x-3)$$. - Vertex at $$x = \frac{5}{2} = 2.5$$. - Calculate vertex y: $$y = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25$$. - Region is below this parabola. **Inequality 4:** $$y \leq -x^2 + 5x + 4$$ - Parabola opens downward. - Vertex at $$x = -\frac{b}{2a} = -\frac{5}{2(-1)} = 2.5$$. - Calculate vertex y: $$y = -(2.5)^2 + 5(2.5) + 4 = -6.25 + 12.5 + 4 = 10.25$$. - Region is on or below this parabola. **Inequality 5:** $$3x - 4y \geq 12$$ - Rewrite in slope-intercept form: $$-4y \geq 12 - 3x$$ $$y \leq \frac{3x}{4} - 3$$ - Region is below or on the line $$y = \frac{3}{4}x - 3$$. 4. **Summary of graphing steps:** - Plot vertices and intercepts for each parabola. - Draw each parabola with correct opening and shading according to inequality. - Plot the line $$y = \frac{3}{4}x - 3$$ and shade below. 5. **Final answer:** The solution is the region satisfying all five inequalities simultaneously, which can be found by graphing each and identifying the overlapping shaded region. This completes the analysis and graphing procedure for the given inequalities.