1. Problem ii: Find the coordinates of the minimum point of the graph of $$f(x)=x^2-4x+2$$. 2. To find the minimum point, we use the vertex formula for a parabola given by $$y=ax^2+bx+c$$. The x-coordinate of the vertex is $$x=-\frac{b}{2a}$$. 3. Here, $$a=1$$ and $$b=-4$$, so $$x=-\frac{-4}{2 \times 1}=\frac{4}{2}=2$$. 4. To find the y-coordinate, substitute $$x=2$$ into $$f(x)$$: $$f(2)=2^2 - 4 \times 2 + 2 = 4 - 8 + 2 = -2$$. 5. So, the minimum point is at $$(2,-2)$$. 6. Problem iii: Sketch the graph of $$f(x)=x^2-4x+2$$, including intercepts and line of symmetry. 7. The y-intercept is found by evaluating $$f(0)$$: $$f(0)=0 - 0 + 2 = 2$$, so y-intercept is $$(0,2)$$. 8. The x-intercepts solve $$x^2 - 4x + 2 = 0$$. 9. Using quadratic formula $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=1,b=-4,c=2$$: $$x=\frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$$. 10. So, x-intercepts at $$\left(2-\sqrt{2},0\right)$$ and $$\left(2+\sqrt{2},0\right)$$. 11. The axis of symmetry is the vertical line $$x=2$$, passing through the vertex. 12. Problem c: Find intersection points between $$y = x^2 - 4x + 2$$ and $$y = -x^2 - 8x$$. 13. Set equal: $$x^2 - 4x + 2 = -x^2 - 8x$$. 14. Move all terms to one side: $$x^2 - 4x + 2 + x^2 + 8x = 0$$ $$2x^2 + 4x + 2 = 0$$. 15. Divide by 2: $$x^2 + 2x + 1 = 0$$. 16. Factor: $$(x+1)^2 = 0$$. 17. So, $$x = -1$$. 18. Find corresponding y: $$y = (-1)^2 - 4(-1) + 2 = 1 + 4 + 2 = 7$$. 19. The graphs intersect at $$(-1,7)$$. Final answers: - Minimum point of $$f(x)$$: $$(2,-2)$$. - Intercepts: y-intercept $$(0,2)$$, x-intercepts $$\left(2-\sqrt{2},0\right)$$ and $$\left(2+\sqrt{2},0\right)$$. - Axis of symmetry: $$x=2$$. - Intersection of graphs: $$(-1,7)$$.