1. The problem involves analyzing three quadratic functions: $$f(x) = -3x^{2} + 4x + 1$$ $$f(x) = 4x^{2} + 4x + 5$$ $$f(x) = x^{2} + 2x + 1$$ We want to understand their graphs, especially the vertex of each parabola. 2. The vertex of a parabola given by $$f(x) = ax^{2} + bx + c$$ is found using the formula for the x-coordinate: $$x = -\frac{b}{2a}$$ The y-coordinate is then: $$y = f\left(-\frac{b}{2a}\right)$$ 3. For the first function $$f(x) = -3x^{2} + 4x + 1$$: - Here, $$a = -3$$, $$b = 4$$, $$c = 1$$. - Calculate $$x$$ of vertex: $$x = -\frac{4}{2 \times -3} = -\frac{4}{-6} = \frac{2}{3}$$ - Calculate $$y$$ of vertex: $$y = -3\left(\frac{2}{3}\right)^{2} + 4\left(\frac{2}{3}\right) + 1 = -3\left(\frac{4}{9}\right) + \frac{8}{3} + 1 = -\frac{12}{9} + \frac{8}{3} + 1 = -\frac{4}{3} + \frac{8}{3} + 1 = \frac{4}{3} + 1 = \frac{7}{3}$$ So the vertex is at $$\left(\frac{2}{3}, \frac{7}{3}\right)$$. 4. For the second function $$f(x) = 4x^{2} + 4x + 5$$: - Here, $$a = 4$$, $$b = 4$$, $$c = 5$$. - Calculate $$x$$ of vertex: $$x = -\frac{4}{2 \times 4} = -\frac{4}{8} = -\frac{1}{2}$$ - Calculate $$y$$ of vertex: $$y = 4\left(-\frac{1}{2}\right)^{2} + 4\left(-\frac{1}{2}\right) + 5 = 4\left(\frac{1}{4}\right) - 2 + 5 = 1 - 2 + 5 = 4$$ So the vertex is at $$\left(-\frac{1}{2}, 4\right)$$. 5. For the third function $$f(x) = x^{2} + 2x + 1$$: - Here, $$a = 1$$, $$b = 2$$, $$c = 1$$. - Calculate $$x$$ of vertex: $$x = -\frac{2}{2 \times 1} = -\frac{2}{2} = -1$$ - Calculate $$y$$ of vertex: $$y = (-1)^{2} + 2(-1) + 1 = 1 - 2 + 1 = 0$$ So the vertex is at $$(-1, 0)$$. 6. Summary: - The vertex of $$f(x) = -3x^{2} + 4x + 1$$ is at $$\left(\frac{2}{3}, \frac{7}{3}\right)$$. - The vertex of $$f(x) = 4x^{2} + 4x + 5$$ is at $$\left(-\frac{1}{2}, 4\right)$$. - The vertex of $$f(x) = x^{2} + 2x + 1$$ is at $$(-1, 0)$$. These vertices represent the highest or lowest points on their respective parabolas depending on the sign of $$a$$ (negative $$a$$ means maximum, positive $$a$$ means minimum).