To sketch the graph of the quadratic equation $y = x^2 + 6x + 8$, let's analyze the equation and find its intercepts with the axes:

1. Finding the y-intercept:

   • The y-intercept occurs when $x = 0$.
   • Substitute $x = 0$ into the equation: $y = 0^2 + 6 \cdot 0 + 8 = 8$
   • So, the y-intercept is at $(0, 8)$.

2. Finding the x-intercepts:

   • The x-intercepts occur when $y = 0$.
   • Set the equation equal to zero: $x^2 + 6x + 8 = 0$
   • Factor the quadratic expression: $(x + 4)(x + 2) = 0$
   • Solve for $x$: $x + 4 = 0 \Rightarrow x = -4$ $x + 2 = 0 \Rightarrow x = -2$
   • So, the x-intercepts are at $(-4, 0)$ and $(-2, 0)$.

3. Sketching the Graph:

   • The equation $y = x^2 + 6x + 8$ represents a parabola that opens upward because the coefficient of $x^2$ is positive.
   • The vertex of the parabola can be found by using the formula $x = -\frac{b}{2a}$: $x = -\frac{6}{2 \cdot 1} = -3$
   • Substitute $x = -3$ back into the equation to find the y-coordinate of the vertex: $y = (-3)^2 + 6 \cdot (-3) + 8 = 9 - 18 + 8 = -1$
   • Thus, the vertex is at $(-3, -1)$.

With this information:

• The y-intercept is at $(0, 8)$.
• The x-intercepts are at $(-4, 0)$ and $(-2, 0)$.
• The vertex is at $(-3, -1)$.

If you'd like, I can provide a sketch based on this analysis or further details on the steps taken.

Would you like a graph or further details?

Here are five questions to explore related concepts:

1. How do we determine the direction in which a parabola opens?
2. What is the significance of the vertex in the graph of a quadratic function?
3. How can the discriminant help in identifying the number of x-intercepts?
4. What changes if the quadratic equation has a negative leading coefficient?
5. How does the axis of symmetry relate to the graph of a quadratic equation?

Tip: The vertex form of a quadratic function can be useful for identifying the vertex directly, making graphing easier.