1. **Problem Statement:** Graph the quadratic functions: a) $y = -(x - 3)^2 + 4$ b) $y = -(x + 1)(x - 1)$ c) $y = 2x^2 - 5x + 2$ 2. **Formula and Rules:** Quadratic functions are of the form $y = ax^2 + bx + c$ or factored/vertex form. Key points to graph: vertex, axis of symmetry, intercepts. 3. **Part a) $y = -(x - 3)^2 + 4$** - This is in vertex form $y = a(x-h)^2 + k$ with vertex $(h,k) = (3,4)$. - Since $a = -1 < 0$, parabola opens downward. - Vertex: $(3,4)$ - Axis of symmetry: $x=3$ - Find y-intercept: set $x=0$, $y = -(0-3)^2 + 4 = -9 + 4 = -5$ - Find x-intercepts: set $y=0$, solve $0 = -(x-3)^2 + 4 \\ (x-3)^2 = 4 \\ x-3 = \pm 2 \\ x = 3 \pm 2$, so $x=1$ or $x=5$ 4. **Part b) $y = -(x + 1)(x - 1)$** - Expand: $y = -(x^2 - 1) = -x^2 + 1$ - Standard form: $a = -1$, $b=0$, $c=1$ - Vertex: $x = -\frac{b}{2a} = 0$, $y = -0^2 + 1 = 1$, so vertex at $(0,1)$ - Parabola opens downward since $a=-1<0$ - x-intercepts: set $y=0$, $0 = -(x+1)(x-1)$, so $x=-1$ or $x=1$ - y-intercept: set $x=0$, $y=1$ 5. **Part c) $y = 2x^2 - 5x + 2$** - Standard form with $a=2$, $b=-5$, $c=2$ - Vertex: $x = -\frac{b}{2a} = -\frac{-5}{4} = \frac{5}{4} = 1.25$ - Calculate $y$ at vertex: $y = 2(1.25)^2 - 5(1.25) + 2 = 2(1.5625) - 6.25 + 2 = 3.125 - 6.25 + 2 = -1.125$ - Parabola opens upward since $a=2>0$ - Find x-intercepts using quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4}$$ - So $x = \frac{5+3}{4} = 2$ or $x = \frac{5-3}{4} = 0.5$ - y-intercept: set $x=0$, $y=2$ **Final answers:** - a) Vertex $(3,4)$, x-intercepts $(1,0)$ and $(5,0)$, y-intercept $(0,-5)$, opens downward. - b) Vertex $(0,1)$, x-intercepts $(-1,0)$ and $(1,0)$, y-intercept $(0,1)$, opens downward. - c) Vertex $(1.25,-1.125)$, x-intercepts $(0.5,0)$ and $(2,0)$, y-intercept $(0,2)$, opens upward. These points can be plotted on the given coordinate planes to complete the graphs.