1. **Problem statement:** Find the turning point and its nature, and the intersection points with the axes for the quadratic function $$y = x^2 - 4x + 3$$. 2. **Turning point:** The turning point of a parabola given by $$y = ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. 3. **Calculate turning point x-coordinate:** Here, $$a = 1$$ and $$b = -4$$, so $$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$$. 4. **Calculate turning point y-coordinate:** Substitute $$x = 2$$ into the function: $$y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$$. 5. **Turning point coordinates:** The turning point is at $$(2, -1)$$. 6. **Nature of turning point:** Since $$a = 1 > 0$$, the parabola opens upwards, so the turning point is a minimum. 7. **Intersection with y-axis:** Set $$x = 0$$: $$y = 0^2 - 4(0) + 3 = 3$$. So the y-intercept is $$(0, 3)$$. 8. **Intersection with x-axis:** Set $$y = 0$$ and solve for $$x$$: $$x^2 - 4x + 3 = 0$$. 9. **Factorize quadratic:** $$x^2 - 4x + 3 = (x - 1)(x - 3) = 0$$. 10. **Solve for x:** $$x - 1 = 0 \Rightarrow x = 1$$ $$x - 3 = 0 \Rightarrow x = 3$$ 11. **x-intercepts:** $$(1, 0)$$ and $$(3, 0)$$. **Final answers:** - Turning point: $$(2, -1)$$ (minimum) - y-intercept: $$(0, 3)$$ - x-intercepts: $$(1, 0)$$ and $$(3, 0)$$