https://file.aiquickdraw.com/mathbot/file/1726127682866wk1zhkdv.jpg

Let's analyze each equation to identify the type of graph they represent.

### (a) \( x^2 - x + y^2 + 3y - \frac{3}{2} = 0 \)

This equation contains both \(x^2\) and \(y^2\), which suggests it might be a circle or an ellipse. Let’s rewrite it to complete the square:

1. Rearrange terms for \(x\) and \(y\):
   \[
   (x^2 - x) + (y^2 + 3y) = \frac{3}{2}
   \]

2. Complete the square for both \(x\) and \(y\):
   - For \(x^2 - x\): Add and subtract \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\)
   - For \(y^2 + 3y\): Add and subtract \(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\)

   \[
   (x - \frac{1}{2})^2 - \frac{1}{4} + (y + \frac{3}{2})^2 - \frac{9}{4} = \frac{3}{2}
   \]

3. Simplify the equation:
   \[
   (x - \frac{1}{2})^2 + (y + \frac{3}{2})^2 = \frac{25}{4}
   \]

This is the equation of a **circle** with center \(\left(\frac{1}{2}, -\frac{3}{2}\right)\) and radius \(\frac{5}{2}\).

### (b) \( x^2 + 4x - 14y = 52 \)

This equation contains \(x^2\), but not \(y^2\), suggesting it is a **parabola**.

1. Rearrange it to get \(x^2\) terms together:
   \[
   x^2 + 4x = 14y + 52
   \]

2. Complete the square for \(x\):
   - Add and subtract \(\left(\frac{4}{2}\right)^2 = 4\)

   \[
   (x + 2)^2 - 4 = 14y + 52
   \]

3. Rearrange:
   \[
   (x + 2)^2 = 14y + 56
   \]
   \[
   (x + 2)^2 = 14(y + 4)
   \]

This is the equation of a **parabola** that opens vertically.

### (c) \( 3x^2 - 42x - 4y^2 - 24y + 99 = 0 \)

This equation contains both \(x^2\) and \(y^2\), but their coefficients have opposite signs, indicating it's likely a **hyperbola**.

1. Group terms:
   \[
   3(x^2 - 14x) - 4(y^2 + 6y) = -99
   \]

2. Complete the square:
   - For \(x^2 - 14x\): Add and subtract \(\left(\frac{14}{2}\right)^2 = 49\)
   - For \(y^2 + 6y\): Add and subtract \(\left(\frac{6}{2}\right)^2 = 9\)

   \[
   3(x - 7)^2 - 147 - 4(y + 3)^2 + 36 = -99
   \]

3. Simplify:
   \[
   3(x - 7)^2 - 4(y + 3)^2 = 12
   \]

4. Divide through by 12:
   \[
   \frac{(x - 7)^2}{4} - \frac{(y + 3)^2}{3} = 1
   \]

This is the equation of a **hyperbola** centered at \((7, -3)\).

### (d) \( 7x^2 - 112x + 2y^2 + 448 = 0 \)

This equation contains both \(x^2\) and \(y^2\) with the same sign, suggesting it is an **ellipse**.

1. Group terms:
   \[
   7(x^2 - 16x) + 2y^2 = -448
   \]

2. Complete the square for \(x\):
   - For \(x^2 - 16x\): Add and subtract \(\left(\frac{16}{2}\right)^2 = 64\)

   \[
   7((x - 8)^2 - 64) + 2y^2 = -448
   \]
   \[
   7(x - 8)^2 - 448 + 2y^2 = -448
   \]

3. Simplify:
   \[
   7(x - 8)^2 + 2y^2 = 0
   \]

This equation does not describe a valid graph since the right-hand side is 0 and both squared terms are positive. However, this might represent a **degenerate conic**, potentially collapsing to a point or line.

### Conclusion:
- (a) **Circle**
- (b) **Parabola**
- (c) **Hyperbola**
- (d) **Degenerate conic** (possibly a collapsed ellipse)

Would you like further clarification on any of these points or additional details?

---

**Related questions:**
1. How do you complete the square for any general quadratic equation?
2. What are the conditions that distinguish ellipses, hyperbolas, and parabolas?
3. Can you explain the geometric interpretation of degenerate conics?
4. What transformations can be applied to conic sections to simplify their equations?
5. How does the center of a conic section influence its equation?

**Tip:** The discriminant of a quadratic equation can also be used to quickly determine the type of conic section represented by the equation.

Identify the graph of each of the following equations: (a) x^2 - x + y^2 + 3y - 3/2 = 0, (b) x^2 + 4x - 14y = 52, (c) 3x^2 - 42x - 4y^2 - 24y + 99 = 0, (d) 7x^2 - 112x + 2y^2 + 448 = 0.

Identify the graphs of conic section equations including circles, parabolas, ellipses, and hyperbolas. Step-by-step solution using algebraic techniques like completing the square. Perfect for Grades 10-12.

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