1. **State the problem:** Given the quadratic function $f(x) = -5x^2 + 11x + 5$, we need to find the vertex, determine if the graph opens up or down, find the axis of symmetry, and find the y-intercept. 2. **Find the vertex using the vertex formula:** The vertex $x$-coordinate is given by $x = -\frac{b}{2a}$ where $a = -5$ and $b = 11$. $$x = -\frac{11}{2 \times -5} = -\frac{11}{-10} = \frac{11}{10}$$ 3. **Find the $y$-coordinate of the vertex:** Substitute $x = \frac{11}{10}$ into $f(x)$: $$f\left(\frac{11}{10}\right) = -5\left(\frac{11}{10}\right)^2 + 11\left(\frac{11}{10}\right) + 5$$ Calculate step-by-step: $$= -5 \times \frac{121}{100} + \frac{121}{10} + 5 = -\frac{605}{100} + \frac{121}{10} + 5$$ Convert all to a common denominator 20: $$-\frac{605}{100} = -\frac{121}{20}, \quad \frac{121}{10} = \frac{242}{20}, \quad 5 = \frac{100}{20}$$ Sum: $$-\frac{121}{20} + \frac{242}{20} + \frac{100}{20} = \frac{-121 + 242 + 100}{20} = \frac{221}{20}$$ So the vertex is $\left(\frac{11}{10}, \frac{221}{20}\right)$. 4. **Determine if the graph opens up or down:** Since $a = -5 < 0$, the parabola opens down. 5. **Find the axis of symmetry:** The axis of symmetry is the vertical line through the vertex $x$-coordinate: $$x = \frac{11}{10}$$ 6. **Find the y-intercept:** The y-intercept occurs at $x=0$: $$f(0) = -5(0)^2 + 11(0) + 5 = 5$$ So the y-intercept is 5. **Final answers:** - Vertex: $\left(\frac{11}{10}, \frac{221}{20}\right)$ - Opens: Down - Axis of symmetry: $x = \frac{11}{10}$ - Y-intercept: 5