1. **State the problem:** Find the minimum point of the graph of $f(x) = x^2 - 4x + 9$. 2. To find the minimum point of a quadratic function $f(x) = ax^2 + bx + c$ (with $a>0$), use the vertex formula: $$x = -\frac{b}{2a}$$ 3. Here, $a = 1$, $b = -4$, so $$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$$ 4. Find $f(2)$: $$f(2) = 2^2 - 4 \times 2 + 9 = 4 - 8 + 9 = 5$$ 5. The minimum point is at $(2, 5)$. --- 6. **Sketch the graph of** $f(x) = x^2 - 4x + 9$. 7. Find the y-intercept by setting $x = 0$: $$f(0) = 0^2 - 4 \times 0 + 9 = 9$$ So, the graph crosses the y-axis at $(0, 9)$. 8. Find the x-intercepts by solving $x^2 - 4x + 9 = 0$: The discriminant is $\Delta = (-4)^2 - 4 \times 1 \times 9 = 16 - 36 = -20 < 0$, so no real roots. Therefore, no x-intercepts. 9. The line of symmetry is the vertical line through the vertex: $$x = 2$$ --- 10. **Find points of intersection between** $$y = x^2 - 4x + 2$$ and $$y = -x^2 - 8x$$ 11. Set the two expressions equal: $$x^2 - 4x + 2 = -x^2 - 8x$$ 12. Bring all terms to one side: $$x^2 - 4x + 2 + x^2 + 8x = 0 \\ 2x^2 + 4x + 2 = 0$$ 13. Divide through by 2: $$x^2 + 2x + 1 = 0$$ 14. This factors as $$(x + 1)^2 = 0$$ 15. So, $x = -1$ is the single solution. 16. Find corresponding $y$: $$y = (-1)^2 - 4(-1) + 2 = 1 + 4 + 2 = 7$$ 17. The two graphs intersect at the point $(-1, 7)$.