1. **Problem:** Solve and analyze the quadratic function $y = x^2 - 4x + 3$. 2. **Formula:** The quadratic function is $y = ax^2 + bx + c$. 3. **Step 1:** Find the vertex using $x = -\frac{b}{2a}$. For $y = x^2 - 4x + 3$, $a=1$, $b=-4$, $c=3$. $$x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$$ 4. **Step 2:** Find $y$ at $x=2$. $$y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$$ Vertex is at $(2, -1)$. 5. **Step 3:** Factor the quadratic. $$y = x^2 - 4x + 3 = (x - 3)(x - 1)$$ 6. **Step 4:** Find roots by setting $y=0$. $$x - 3 = 0 \Rightarrow x=3$$ $$x - 1 = 0 \Rightarrow x=1$$ 7. **Step 5:** Axis of symmetry is $x=2$. 8. **Step 6:** Since $a=1 > 0$, parabola opens upward. --- 1. **Problem:** Solve and analyze $y = -2x^2 + 8x - 5$. 2. $a=-2$, $b=8$, $c=-5$. 3. Vertex $x = -\frac{8}{2 \times -2} = -\frac{8}{-4} = 2$. 4. $y = -2(2)^2 + 8(2) - 5 = -8 + 16 - 5 = 3$. Vertex at $(2, 3)$. 5. Find roots: $$-2x^2 + 8x - 5 = 0$$ Divide both sides by $-1$: $$\cancel{-}2x^2 + \cancel{8}x - \cancel{5} = 0 \Rightarrow 2x^2 - 8x + 5 = 0$$ Use quadratic formula: $$x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 2 \times 5}}{2 \times 2} = \frac{8 \pm \sqrt{64 - 40}}{4} = \frac{8 \pm \sqrt{24}}{4}$$ Simplify $\sqrt{24} = 2\sqrt{6}$: $$x = \frac{8 \pm 2\sqrt{6}}{4} = \frac{\cancel{2} \times 4 \pm \cancel{2} \sqrt{6}}{\cancel{2} \times 2} = \frac{4 \pm \sqrt{6}}{2}$$ 6. Axis of symmetry $x=2$. 7. Since $a=-2 < 0$, parabola opens downward. --- 1. **Problem:** Solve and analyze $y = \frac{1}{2}x^2 + x - 4$. 2. $a=\frac{1}{2}$, $b=1$, $c=-4$. 3. Vertex $x = -\frac{1}{2 \times \frac{1}{2}} = -\frac{1}{1} = -1$. 4. $y = \frac{1}{2}(-1)^2 + (-1) - 4 = \frac{1}{2} - 1 - 4 = -\frac{9}{2} = -4.5$. Vertex at $(-1, -4.5)$. 5. Find roots: $$\frac{1}{2}x^2 + x - 4 = 0$$ Multiply both sides by 2: $$\cancel{2} \times \frac{1}{2}x^2 + \cancel{2} \times x - \cancel{2} \times 4 = 0 \Rightarrow x^2 + 2x - 8 = 0$$ Factor: $$(x + 4)(x - 2) = 0$$ Roots: $x = -4$, $x = 2$. 6. Axis of symmetry $x = -1$. 7. Since $a=\frac{1}{2} > 0$, parabola opens upward. --- 1. **Problem:** Solve and analyze $y = x^2 - 6x + 9$. 2. $a=1$, $b=-6$, $c=9$. 3. Vertex $x = -\frac{-6}{2 \times 1} = \frac{6}{2} = 3$. 4. $y = (3)^2 - 6(3) + 9 = 9 - 18 + 9 = 0$. Vertex at $(3, 0)$. 5. Factor: $$y = (x - 3)^2$$ 6. Root at $x=3$ (double root). 7. Axis of symmetry $x=3$. 8. Since $a=1 > 0$, parabola opens upward. --- 1. **Problem:** Solve and analyze $y = -x^2 + 2x + 1$. 2. $a=-1$, $b=2$, $c=1$. 3. Vertex $x = -\frac{2}{2 \times -1} = -\frac{2}{-2} = 1$. 4. $y = -(1)^2 + 2(1) + 1 = -1 + 2 + 1 = 2$. Vertex at $(1, 2)$. 5. Find roots: $$-x^2 + 2x + 1 = 0$$ Multiply both sides by $-1$: $$\cancel{-}x^2 + \cancel{2}x + \cancel{1} = 0 \Rightarrow x^2 - 2x - 1 = 0$$ Use quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-1)}}{2} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2}$$ Simplify $\sqrt{8} = 2\sqrt{2}$: $$x = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$$ 6. Axis of symmetry $x=1$. 7. Since $a=-1 < 0$, parabola opens downward.