1. Problem statement: Sketch the graph of the quadratic function $f(x) = x^2 - 6x + 8$ and find the axis of symmetry. 2. To graph $f(x) = x^2 - 6x + 8$, first find its vertex by completing the square or using the vertex formula. The vertex formula for $x$-coordinate is $x = \frac{-b}{2a}$. 3. Here, $a = 1$ and $b = -6$, so: $$ x = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3 $$ 4. Substitute $x=3$ into $f(x)$ to find the $y$-coordinate of the vertex: $$ f(3) = 3^2 - 6(3) + 8 = 9 - 18 + 8 = -1 $$ 5. The vertex is at $(3, -1)$. This is the minimum point because $a>0$. 6. Find the $y$-intercept by substituting $x=0$: $$ f(0) = 0^2 - 6(0) + 8 = 8 $$ 7. Find the $x$-intercepts by setting $f(x)=0$ and solving: $$ x^2 - 6x + 8 = 0 $$ 8. Factor the quadratic: $$ (x-2)(x-4) = 0 $$ 9. So, $$ x = 2 \text{ or } x = 4 $$ 10. The axis of symmetry is the vertical line through the vertex at $x=3$. Final answers: - The graph is a parabola opening upward with vertex at $(3, -1)$. - Intercepts are at $x=2$, $x=4$, and $y=8$. - Axis of symmetry: $$ x = 3 $$