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The problem is asking to write the equations of the parabolas in standard form and identify key components such as the vertex, focus, directrix, axis of symmetry (AOS), the direction of opening, and to sketch the graphs. Let's solve each equation step by step.

### Problem 7: \( y^2 + 4x + 8y = -4 \)
We aim to rewrite this equation in standard form, which is either:
- \( (y - k)^2 = 4p(x - h) \) (horizontal parabola), or
- \( (x - h)^2 = 4p(y - k) \) (vertical parabola).

This equation involves \( y^2 \), so it represents a **horizontal** parabola. Let's complete the square for \( y \) to get the standard form.

1. **Rewrite the equation:**
   \[
   y^2 + 8y + 4x = -4
   \]

2. **Complete the square on \( y \):**
   The coefficient of \( y \) is 8, so half of 8 is 4, and squaring it gives \( 16 \). Add and subtract 16:
   \[
   (y^2 + 8y + 16) + 4x = -4 + 16
   \]
   This simplifies to:
   \[
   (y + 4)^2 + 4x = 12
   \]
   Now isolate \( x \):
   \[
   4x = 12 - (y + 4)^2
   \]
   \[
   x = 3 - \frac{1}{4}(y + 4)^2
   \]
   This is the standard form \( (y - k)^2 = 4p(x - h) \), with \( h = 3 \), \( k = -4 \), and \( p = -\frac{1}{4} \).

Now we can identify the required components:
- **Vertex**: \( (3, -4) \)
- **Focus**: Since \( 4p = -1 \), \( p = -\frac{1}{4} \). The focus is \( \left(3 - \frac{1}{4}, -4 \right) = \left( \frac{11}{4}, -4 \right) \).
- **Directrix**: The directrix is the vertical line \( x = 3 + \frac{1}{4} = \frac{13}{4} \).
- **AOS (Axis of Symmetry)**: The axis of symmetry is \( y = -4 \).
- **Opening**: The parabola opens to the **left** because \( p < 0 \).

### Problem 8: \( x^2 + 6x + 8y = 7 \)
We aim to rewrite this equation in standard form for a **vertical parabola** because it involves \( x^2 \). Let's complete the square for \( x \).

1. **Rewrite the equation:**
   \[
   x^2 + 6x + 8y = 7
   \]

2. **Complete the square on \( x \):**
   The coefficient of \( x \) is 6, so half of 6 is 3, and squaring it gives \( 9 \). Add and subtract 9:
   \[
   (x^2 + 6x + 9) + 8y = 7 + 9
   \]
   This simplifies to:
   \[
   (x + 3)^2 + 8y = 16
   \]
   Now isolate \( y \):
   \[
   8y = 16 - (x + 3)^2
   \]
   \[
   y = 2 - \frac{1}{8}(x + 3)^2
   \]
   This is the standard form \( (x - h)^2 = 4p(y - k) \), with \( h = -3 \), \( k = 2 \), and \( p = -\frac{1}{8} \).

Now we can identify the required components:
- **Vertex**: \( (-3, 2) \)
- **Focus**: Since \( 4p = -1 \), \( p = -\frac{1}{8} \). The focus is \( \left( -3, 2 - \frac{1}{8} \right) = \left( -3, \frac{15}{8} \right) \).
- **Directrix**: The directrix is the line \( y = 2 + \frac{1}{8} = \frac{17}{8} \).
- **AOS (Axis of Symmetry)**: The axis of symmetry is \( x = -3 \).
- **Opening**: The parabola opens **downward** because \( p < 0 \).

Would you like a detailed graph of these solutions, or do you have any specific questions?

### 5 Related Questions:
1. How do you determine if a parabola opens horizontally or vertically?
2. What are the steps for completing the square for a quadratic equation?
3. How does the value of \( p \) affect the focus and directrix of a parabola?
4. What is the significance of the axis of symmetry in a parabola?
5. Can you explain the relationship between the focus, directrix, and vertex of a parabola?

### Tip:
Always ensure that the squared term in the equation of a parabola (whether \( x^2 \) or \( y^2 \)) dictates the direction in which it opens—horizontal for \( y^2 \) and vertical for \( x^2 \).

Write the equation of the parabola in standard form and identify the a) vertex, b) focus, c) directrix, d) AOS, e) opening of the parabola, and f) sketch of the graph for the following equations: 7. y^2 + 4x + 8y = -4, 8. x^2 + 6x + 8y = 7

Learn how to convert parabolas into standard form, identify their vertices, foci, directrices, axes of symmetry, and direction of opening for equations like y^2 + 4x + 8y = -4 and x^2 + 6x + 8y = 7. This step-by-step guide is perfect for high school students studying conic sections.

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