Vertex Parabola Eaef28

1. **State the problem:** Find the vertex of the parabola given by the equation $$y = - \frac{4}{9} x^2 + \frac{8}{3} x$$ using the method of completing the square. 2. **Recall the formula and rules:** The vertex form of a quadratic is $$y = a(x-h)^2 + k$$ where $ (h,k) $ is the vertex. Completing the square helps rewrite the quadratic in this form. 3. **Factor out the common factor from the quadratic terms:** $$y = - \frac{4}{9} \left(x^2 - \frac{8}{3} \div \left(- \frac{4}{9}\right) x \right)$$ Calculate the factor inside: $$\frac{8}{3} \div \left(- \frac{4}{9}\right) = \frac{8}{3} \times \left(- \frac{9}{4}\right) = -6$$ So, $$y = - \frac{4}{9} (x^2 - 6x)$$ 4. **Complete the square inside the parentheses:** Take half of the coefficient of $x$, which is $-6$, half is $-3$, square it: $(-3)^2 = 9$. Add and subtract $9$ inside the parentheses: $$y = - \frac{4}{9} (x^2 - 6x + 9 - 9)$$ 5. **Rewrite as a perfect square and simplify:** $$y = - \frac{4}{9} \left((x - 3)^2 - 9\right)$$ Distribute: $$y = - \frac{4}{9} (x - 3)^2 + \frac{4}{9} \times 9$$ Simplify: $$y = - \frac{4}{9} (x - 3)^2 + 4$$ 6. **Identify the vertex:** The vertex form is $$y = a(x-h)^2 + k$$ with $a = - \frac{4}{9}$, $h = 3$, and $k = 4$. So the vertex is at $ (3, 4) $. 7. **Determine if maximum or minimum:** Since $a = - \frac{4}{9} < 0$, the parabola opens downward, so the vertex is a maximum point. **Final answer:** The vertex is at $ (3, 4) $, which is the maximum point of the parabola.