1. Problem 1: For the function $f(x) = -x^2 - 2x - 3$ - **Vertex**: Use vertex formula $x = -\frac{b}{2a} = -\frac{-2}{2(-1)} = 1$ - Substitute $x=1$ to find $y$: $f(1) = -(1)^2 - 2(1) - 3 = -1 - 2 - 3 = -6$ - Vertex is $(1, -6)$ - **Opening**: Since $a = -1 < 0$, the parabola opens downward. - **Vertex point type**: It's a maximum point as parabola opens downward. - **Axis of symmetry** equation: $x=1$ - **Domain**: All real numbers, written $(-\infty, \infty)$. - **Range**: Since vertex is max $y=-6$, range is $(-\infty, -6]$. 2. Problem 2: For $f(x) = 2x^2 + 4x - 3$ - Vertex formula: $x = -\frac{4}{2(2)} = -1$ - $f(-1) = 2(-1)^2 + 4(-1) - 3 = 2 - 4 - 3 = -5$ - Vertex: $(-1, -5)$ - Opening: $a = 2 > 0$ so opens upward - Vertex point type: minimum - Axis of symmetry: $x = -1$ - Domain: $(-\infty, \infty)$ - Range: $[-5, \infty)$ 3. Problem 3: $f(x) = (x+2)^2 + 3$ - Vertex form shows vertex directly: $(-2, 3)$ - Opening: $a=1 > 0$ opens upward - Vertex type: minimum - Domain: $(-\infty, \infty)$ - Range: $[3, \infty)$ 4. Convert $f(x) = x^2 + 2x - 1$ to vertex form - Complete the square: $$x^2 + 2x = (x+1)^2 - 1$$ - So, $$f(x) = (x+1)^2 - 1 - 1 = (x+1)^2 - 2$$ 5. Convert $f(x) = (x+2)^2 - 4$ to standard form - Expand: $$(x+2)^2 - 4 = x^2 + 4x + 4 - 4 = x^2 + 4x$$ 6. Convert $f(x) = 2(x-10)^2 + 14$ to standard form - Expand: $$2(x-10)^2 + 14 = 2(x^2 - 20x + 100) + 14 = 2x^2 - 40x + 200 + 14 = 2x^2 - 40x + 214$$ Summary: - Vertex points and graph shapes explained. - Axis of symmetry and range/domain detailed. - Conversion between standard and vertex forms demonstrated with full steps.