Inequality Graph A4D343

1. **State the problem:** We need to find the solution region for the system of inequalities: $$y \geq 2(1.3)^x$$ $$y \leq -x^2 + 4x + 5$$ 2. **Understand the inequalities:** - The first inequality $$y \geq 2(1.3)^x$$ represents the region above or on the exponential curve. - The second inequality $$y \leq -x^2 + 4x + 5$$ represents the region below or on the downward parabola. 3. **Analyze the parabola:** - The parabola is $$y = -x^2 + 4x + 5$$. - Its vertex is at $$x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2$$. - Substitute $$x=2$$ to find vertex $$y$$: $$y = -(2)^2 + 4(2) + 5 = -4 + 8 + 5 = 9$$. - So vertex is at $$(2, 9)$$. - Intercepts: - At $$x=0$$, $$y = 5$$. - At $$y=0$$, solve $$0 = -x^2 + 4x + 5$$: $$x^2 - 4x - 5 = 0$$ $$(x - 5)(x + 1) = 0$$ $$x = 5$$ or $$x = -1$$. 4. **Analyze the exponential curve:** - The curve is $$y = 2(1.3)^x$$. - At $$x=0$$, $$y=2$$. - At $$x=6$$, $$y = 2(1.3)^6 \approx 2 \times 6.16 = 12.32$$. - At $$x=-3$$, $$y = 2(1.3)^{-3} = 2 / (1.3)^3 \approx 2 / 2.197 = 0.91$$. 5. **Find intersection points:** - The curves intersect near $$x = -0.8$$ with $$y \approx 1.8$$. 6. **Determine the solution region:** - The solution region is where $$y \geq 2(1.3)^x$$ (above the exponential curve) and $$y \leq -x^2 + 4x + 5$$ (below the parabola). - This region lies between the parabola and the exponential curve. 7. **Match with the graph descriptions:** - The parabola vertex at $$(2, 9)$$ and intercepts at $$(0, 5)$$ and $$(5, 0)$$ matches the parabola equation. - The exponential curve passes through $$(6, 10)$$ and $$(-3, 1)$$ and intercepts the y-axis at 2. - The intersection point near $$(-0.8, 1.8)$$ is consistent. **Final answer:** The correct graph is the one showing a downward parabola with vertex at $$(2, 9)$$, intercepts at $$(0, 5)$$ and $$(5, 0)$$, and an exponential curve passing through $$(6, 10)$$ and $$(-3, 1)$$ with y-intercept 2, intersecting near $$(-0.8, 1.8)$$. This corresponds to the first or third option as described.