1. **State the problem:** We are given four quadratic equations and two graphs with their properties. We need to match each graph to the correct equation and find the axis of symmetry, vertex, domain, range, and zeros. 2. **Recall formulas and rules:** - The axis of symmetry for a quadratic $y = ax^2 + bx + c$ is $x = -\frac{b}{2a}$. - The vertex is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$. - The domain of any quadratic is all real numbers: $(-\infty, \infty)$. - The range depends on the sign of $a$: - If $a > 0$, parabola opens upward, range is $[y_{vertex}, \infty)$. - If $a < 0$, parabola opens downward, range is $(-\infty, y_{vertex}]$. - Zeros are the $x$-intercepts where $y=0$. 3. **Analyze each equation:** A. $y = x^2 + 2x + 1$ - $a=1$, $b=2$, $c=1$ - Axis: $x = -\frac{2}{2\times1} = -1$ - Vertex: $y = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0$ - Domain: $(-\infty, \infty)$ - Range: Since $a=1>0$, range is $[0, \infty)$ - Zeros: Solve $x^2 + 2x + 1 = 0$ which factors as $(x+1)^2=0$, zero at $x=-1$ B. $y = -x^2 + 2x + 1$ - $a=-1$, $b=2$, $c=1$ - Axis: $x = -\frac{2}{2\times(-1)} = 1$ - Vertex: $y = -1^2 + 2(1) + 1 = -1 + 2 + 1 = 2$ - Domain: $(-\infty, \infty)$ - Range: Since $a=-1<0$, range is $(-\infty, 2]$ - Zeros: Solve $-x^2 + 2x + 1=0$ or $x^2 - 2x -1=0$ with roots $x=1 \pm \sqrt{2}$ C. $y = x^2 - 2x + 1$ - $a=1$, $b=-2$, $c=1$ - Axis: $x = -\frac{-2}{2\times1} = 1$ - Vertex: $y = 1^2 - 2(1) + 1 = 1 - 2 + 1 = 0$ - Domain: $(-\infty, \infty)$ - Range: $[0, \infty)$ - Zeros: $(x-1)^2=0$, zero at $x=1$ D. $y = -x^2 - 2x + 1$ - $a=-1$, $b=-2$, $c=1$ - Axis: $x = -\frac{-2}{2\times(-1)} = -1$ - Vertex: $y = -(-1)^2 - 2(-1) + 1 = -1 + 2 + 1 = 2$ - Domain: $(-\infty, \infty)$ - Range: $(-\infty, 2]$ - Zeros: Solve $-x^2 - 2x + 1=0$ or $x^2 + 2x -1=0$ roots $x=-1 \pm \sqrt{2}$ 4. **Match graphs:** - Top-right graph: upward parabola, vertex at $(-1,0)$, axis $x=-1$, range $y \geq 0$, zero at $x=-1$ matches equation A. - Bottom-left graph: downward parabola, vertex at $(1,0)$, axis $x=1$, range $y \leq 0$, zero at $x=1$ matches none exactly since vertex $y=0$ but B and D have vertex $y=2$. However, C has vertex at $(1,0)$ but opens upward, so the downward parabola with vertex at $(1,0)$ is not exactly given. The closest is B with vertex at $(1,2)$ but range $y \leq 2$. Since the bottom-left graph has vertex at $(1,0)$ and opens downward, none of the given equations exactly match that vertex value. Possibly a typo in the problem. Assuming the vertex is at $(1,0)$ and downward opening, the equation would be $y = - (x-1)^2$ which is $y = -x^2 + 2x -1$, not listed. 5. **Final answers for top-right graph (A):** - Axis of Symmetry: $x = -1$ - Vertex: $(-1, 0)$ - Domain: $(-\infty, \infty)$ - Range: $[0, \infty)$ - Zeros: $x = -1$ - Equation: $y = x^2 + 2x + 1$ For bottom-left graph, closest match is equation C but it opens upward. If we accept vertex at $(1,0)$ and upward opening: - Axis of Symmetry: $x=1$ - Vertex: $(1,0)$ - Domain: $(-\infty, \infty)$ - Range: $[0, \infty)$ - Zeros: $x=1$ - Equation: $y = x^2 - 2x + 1$ "slug":"parabola-properties","subject":"algebra","svg":"","desmos":{"latex":"y=x^2+2x+1","features":{"intercepts":true,"extrema":true}},"q_count":2