1. The problem asks to match the given graphs to one of the four quadratic functions: A) $f(x) = x^2 + 2x + 1$ B) $f(x) = x^2 + 2x$ C) $f(x) = x^2 - 2x$ D) $f(x) = x^2 - 2x + 1$ 2. Important: The vertex form of a parabola is $f(x) = a(x-h)^2 + k$ where $(h,k)$ is the vertex. 3. Let's find the vertex of each function by completing the square or using vertex formula $h = -\frac{b}{2a}$ for $ax^2 + bx + c$. - For A) $x^2 + 2x + 1$: $$h = -\frac{2}{2 \times 1} = -1$$ $$k = f(-1) = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0$$ Vertex: $(-1,0)$ - For B) $x^2 + 2x$: $$h = -\frac{2}{2} = -1$$ $$k = f(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1$$ Vertex: $(-1,-1)$ - For C) $x^2 - 2x$: $$h = -\frac{-2}{2} = 1$$ $$k = f(1) = 1^2 - 2(1) = 1 - 2 = -1$$ Vertex: $(1,-1)$ - For D) $x^2 - 2x + 1$: $$h = -\frac{-2}{2} = 1$$ $$k = f(1) = 1^2 - 2(1) + 1 = 1 - 2 + 1 = 0$$ Vertex: $(1,0)$ 4. The graphs both have vertex at $(-1,0)$, so the matching functions must have vertex $(-1,0)$. 5. From above, only function A) $x^2 + 2x + 1$ has vertex $(-1,0)$. 6. Therefore, the graph corresponds to function A). Final answer: Function A) $f(x) = x^2 + 2x + 1$ matches the graph with vertex at $(-1,0)$.