1. We are asked to graph each quadratic equation and label the vertex and axis of symmetry. 2. The vertex form of a quadratic is $y = a(x-h)^2 + k$, where $(h,k)$ is the vertex and the axis of symmetry is $x = h$. 3. For standard form $y = ax^2 + bx + c$, the vertex is at $x = -\frac{b}{2a}$ and $y$ is found by substituting this $x$ back into the equation. 4. For factored form $y = a(x-r_1)(x-r_2)$, the axis of symmetry is the midpoint of the roots $r_1$ and $r_2$. --- a) $y = 2(x - 3)^2 - 9$ - Vertex is $(3, -9)$ directly from the form. - Axis of symmetry is $x = 3$. b) $y = -\frac{1}{2}(x + 1)^2 + 3$ - Vertex is $(-1, 3)$. - Axis of symmetry is $x = -1$. c) $y = -x^2 + 4x + 4$ - Find vertex $x = -\frac{b}{2a} = -\frac{4}{2(-1)} = 2$. - Substitute $x=2$: $y = -(2)^2 + 4(2) + 4 = -4 + 8 + 4 = 8$. - Vertex is $(2, 8)$. - Axis of symmetry is $x = 2$. --- Intermediate step showing cancellation: $$x = -\frac{\cancel{4}}{2\times \cancel{-1}} = 2$$ d) $y = \frac{1}{2}x^2 - 4x + 10$ - Vertex $x = -\frac{b}{2a} = -\frac{-4}{2 \times \frac{1}{2}} = \frac{4}{1} = 4$. - Substitute $x=4$: $y = \frac{1}{2}(4)^2 - 4(4) + 10 = \frac{1}{2} \times 16 - 16 + 10 = 8 - 16 + 10 = 2$. - Vertex is $(4, 2)$. - Axis of symmetry is $x = 4$. --- Intermediate step showing cancellation: $$x = -\frac{-4}{2 \times \frac{1}{2}} = -\frac{-4}{1} = 4$$ e) $y = (x - 4)(x + 2)$ - Roots are $4$ and $-2$. - Axis of symmetry is midpoint: $x = \frac{4 + (-2)}{2} = 1$. - Find vertex by substituting $x=1$: $y = (1 - 4)(1 + 2) = (-3)(3) = -9$. - Vertex is $(1, -9)$. f) $y = 2(x + 1)(x + 5)$ - Roots are $-1$ and $-5$. - Axis of symmetry is $x = \frac{-1 + (-5)}{2} = -3$. - Substitute $x = -3$: $y = 2(-3 + 1)(-3 + 5) = 2(-2)(2) = -8$. - Vertex is $(-3, -8)$. --- Final answers: - a) Vertex $(3, -9)$, axis $x=3$ - b) Vertex $(-1, 3)$, axis $x=-1$ - c) Vertex $(2, 8)$, axis $x=2$ - d) Vertex $(4, 2)$, axis $x=4$ - e) Vertex $(1, -9)$, axis $x=1$ - f) Vertex $(-3, -8)$, axis $x=-3$