1. The problem is to sketch the curve given by the quadratic function $$y = x^2 - 6x + 8$$. 2. First, identify the key features of the quadratic: the vertex, axis of symmetry, and intercepts. 3. Find the vertex using the formula $$x = -\frac{b}{2a}$$ for a quadratic $$ax^2 + bx + c$$. Here, $$a = 1$$, $$b = -6$$, and $$c = 8$$. 4. Calculate $$x$$ coordinate of vertex: $$x = -\frac{-6}{2 \times 1} = \frac{6}{2} = 3$$. 5. Find the $$y$$ coordinate by substituting $$x = 3$$ into the function: $$y = 3^2 - 6 \times 3 + 8 = 9 - 18 + 8 = -1$$. 6. So, the vertex is at $$ (3, -1) $$. 7. The axis of symmetry is the vertical line $$x = 3$$. 8. Find the y-intercept by substituting $$x=0$$: $$y = 0 - 0 + 8 = 8$$, so the y-intercept is $$ (0,8) $$. 9. Find x-intercepts by solving $$x^2 - 6x + 8 = 0$$. 10. Factor the quadratic: $$x^2 - 6x + 8 = (x - 2)(x - 4)$$. 11. Set factors equal to zero: $$x - 2 = 0 \Rightarrow x=2$$ and $$x - 4=0 \Rightarrow x=4$$. 12. So, the x-intercepts are at $$ (2,0) $$ and $$ (4,0) $$. 13. The parabola opens upwards since $$a = 1 > 0$$. 14. To sketch: draw the vertex $$ (3,-1) $$, plot intercepts $$ (0,8), (2,0), (4,0) $$, and draw a smooth curve passing through these points with an axis of symmetry at $$x=3$$. 15. Summary: vertex is $$ (3,-1) $$, axis of symmetry $$x=3$$, x-intercepts at $$2$$ and $$4$$, y-intercept at $$8$$, parabola opens upward.