1. **State the problem:** We need to match each quadratic equation to its corresponding graph based on the vertex and the direction the parabola opens. 2. **Recall the vertex formula:** For a quadratic equation $y = ax^2 + bx + c$, the vertex is at $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$$ and the parabola opens upward if $a > 0$ and downward if $a < 0$. 3. **Analyze each equation:** - For $y = -x^2 + 5x - 2$: - $a = -1$, $b = 5$, $c = -2$ - Vertex $x$-coordinate: $$-\frac{5}{2 \times -1} = \frac{5}{2} = 2.5$$ - Vertex $y$-coordinate: $$-(2.5)^2 + 5(2.5) - 2 = -6.25 + 12.5 - 2 = 4.25$$ - Parabola opens downward since $a = -1 < 0$. - For $y = x^2 + 3x + 2$: - $a = 1$, $b = 3$, $c = 2$ - Vertex $x$-coordinate: $$-\frac{3}{2 \times 1} = -1.5$$ - Vertex $y$-coordinate: $$( -1.5 )^2 + 3(-1.5) + 2 = 2.25 - 4.5 + 2 = -0.25$$ - Parabola opens upward since $a = 1 > 0$. - For $y = x^2 - 1$: - $a = 1$, $b = 0$, $c = -1$ - Vertex $x$-coordinate: $$-\frac{0}{2 \times 1} = 0$$ - Vertex $y$-coordinate: $$0^2 - 1 = -1$$ - Parabola opens upward since $a = 1 > 0$. 4. **Match to graphs:** - Left graph: upward-opening parabola with vertex at $(0, -1)$ matches $y = x^2 - 1$. - Middle graph: downward-opening parabola with vertex at $(2.5, 4.25)$ matches $y = -x^2 + 5x - 2$. - Right graph: upward-opening parabola with vertex at $(-1.5, -0.25)$ matches $y = x^2 + 3x + 2$. **Final answer:** - Left graph: $y = x^2 - 1$ - Middle graph: $y = -x^2 + 5x - 2$ - Right graph: $y = x^2 + 3x + 2$