1. The problem is to analyze the function given in the document link, but since the link is inaccessible here, I will demonstrate solving a typical algebra problem involving a quadratic function. 2. Consider the quadratic function $$y = x^2 - 4x + 3$$. 3. The formula to find the roots (x-intercepts) of a quadratic equation $$ax^2 + bx + c = 0$$ is given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 4. For our function, $$a=1$$, $$b=-4$$, and $$c=3$$. 5. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 3 = 16 - 12 = 4$$ 6. Since $$\Delta > 0$$, there are two distinct real roots. 7. Calculate the roots: $$x = \frac{-(-4) \pm \sqrt{4}}{2 \times 1} = \frac{4 \pm 2}{2}$$ 8. Simplify each root: $$x_1 = \frac{4 + 2}{2} = \frac{6}{2} = 3$$ $$x_2 = \frac{4 - 2}{2} = \frac{2}{2} = 1$$ 9. The roots are $$x=3$$ and $$x=1$$, which are the x-intercepts of the parabola. 10. The vertex of the parabola can be found using $$x = -\frac{b}{2a} = -\frac{-4}{2 \times 1} = 2$$. 11. Substitute $$x=2$$ into the function to find the y-coordinate of the vertex: $$y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$$ 12. So, the vertex is at $$(2, -1)$$. 13. The parabola opens upwards since $$a=1 > 0$$. Final answer: The quadratic function $$y = x^2 - 4x + 3$$ has roots at $$x=1$$ and $$x=3$$, and its vertex is at $$(2, -1)$$.