1. **State the problem:** Write the quadratic equation $$y = -\frac{7}{4}x^2 - \frac{1}{2}x + 2$$ in vertex form, and identify the vertex, direction of opening, and y-intercept. 2. **Recall the vertex form of a quadratic:** $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex. 3. **Complete the square:** Start with $$y = -\frac{7}{4}x^2 - \frac{1}{2}x + 2$$ Factor out the coefficient of $x^2$ from the first two terms: $$y = -\frac{7}{4}\left(x^2 + \cancel{\frac{1}{2} \div -\frac{7}{4}}x\right) + 2$$ Calculate the factor inside: $$\frac{-\frac{1}{2}}{-\frac{7}{4}} = \frac{-0.5}{-1.75} = \frac{2}{7}$$ So rewrite: $$y = -\frac{7}{4}\left(x^2 + \frac{2}{7}x\right) + 2$$ 4. **Find the term to complete the square:** Take half of $\frac{2}{7}$ and square it: $$\left(\frac{1}{7}\right)^2 = \frac{1}{49}$$ Add and subtract inside the parentheses: $$y = -\frac{7}{4}\left(x^2 + \frac{2}{7}x + \frac{1}{49} - \frac{1}{49}\right) + 2$$ 5. **Rewrite as a perfect square and simplify:** $$y = -\frac{7}{4}\left(\left(x + \frac{1}{7}\right)^2 - \frac{1}{49}\right) + 2$$ Distribute: $$y = -\frac{7}{4}\left(x + \frac{1}{7}\right)^2 + \frac{7}{4} \times \frac{1}{49} + 2$$ Calculate: $$\frac{7}{4} \times \frac{1}{49} = \frac{7}{196} = \frac{1}{28}$$ So: $$y = -\frac{7}{4}\left(x + \frac{1}{7}\right)^2 + \frac{1}{28} + 2$$ Convert 2 to $\frac{56}{28}$: $$y = -\frac{7}{4}\left(x + \frac{1}{7}\right)^2 + \frac{57}{28}$$ 6. **Identify the vertex:** The vertex is at $$\left(-\frac{1}{7}, \frac{57}{28}\right)$$ 7. **Direction of opening:** Since $a = -\frac{7}{4} < 0$, the parabola opens downwards. 8. **Find the y-intercept:** Set $x=0$: $$y = -\frac{7}{4}(0)^2 - \frac{1}{2}(0) + 2 = 2$$ So the y-intercept is at $(0, 2)$. **Final answer:** $$y = -\frac{7}{4}\left(x + \frac{1}{7}\right)^2 + \frac{57}{28}$$ Vertex: $$\left(-\frac{1}{7}, \frac{57}{28}\right)$$ Direction: Downwards Y-intercept: $(0, 2)$