1. The problem is to graph the quadratic function $y = x^2 + 5x + 6$. 2. The general form of a quadratic function is $y = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 3. For $y = x^2 + 5x + 6$, we have $a=1$, $b=5$, and $c=6$. 4. To find the roots (x-intercepts), solve $x^2 + 5x + 6 = 0$. 5. Factor the quadratic: $x^2 + 5x + 6 = (x + 2)(x + 3) = 0$. 6. Set each factor equal to zero: $x + 2 = 0$ or $x + 3 = 0$. 7. Solve for $x$: $x = -2$ or $x = -3$. 8. The roots are $x = -2$ and $x = -3$, so the graph crosses the x-axis at these points. 9. The y-intercept is found by evaluating $y$ at $x=0$: $y = 0^2 + 5(0) + 6 = 6$. 10. The vertex of the parabola is at $x = -\frac{b}{2a} = -\frac{5}{2(1)} = -\frac{5}{2} = -2.5$. 11. Calculate $y$ at the vertex: $y = (-2.5)^2 + 5(-2.5) + 6 = 6.25 - 12.5 + 6 = -0.25$. 12. The vertex is at $(-2.5, -0.25)$, which is the minimum point since $a > 0$. 13. The parabola opens upwards because $a=1 > 0$. Final answer: The graph of $y = x^2 + 5x + 6$ is a parabola opening upwards with roots at $x = -3$ and $x = -2$, y-intercept at $(0,6)$, and vertex at $(-2.5, -0.25)$.