using these as standard form:
(y-k)^2=4a(x-h)
(y-k)^2=-4a(x-h)
(x-h)^2=4a(y-k)
(x-h)^2=-4a(y-k)

[PROVIDE SOLUTIONS]
Identify the vertex, focus, axis of symmetry, and directrix of each. Then sketch the graph.

1. y=-(x+3)^2)-1
2. x=-(1/4)((y+2)^2)
3. -(1/3)(x-3)=(y+5)^2
4. -(y+2)=(x-2)^2
5. 3y+4x=-(2x^2)-14
6. x=-2(y+2)^2

Let's analyze each equation using the standard forms provided:

### 1. \( y = -(x + 3)^2 - 1 \)

This equation matches the standard form \( (x-h)^2 = -4a(y-k) \), which describes a parabola that opens downward.

- **Vertex:** The vertex \((h, k)\) is given by \((-3, -1)\).
- **Focus:** The standard form is \( (x-h)^2 = -4a(y-k) \), where \( a = \frac{1}{4} \). Here, \( a = \frac{1}{4} \). The focus is at \((h, k - a)\), so \((-3, -1 + \frac{1}{4}) = (-3, -1.25)\).
- **Axis of Symmetry:** The axis of symmetry is the vertical line \( x = -3 \).
- **Directrix:** The directrix is the line \( y = k + a \), so \( y = -1 + \frac{1}{4} = -0.75 \).

### 2. \( x = -\frac{1}{4}(y + 2)^2 \)

This equation matches the standard form \( (y-k)^2 = -4a(x-h) \), which describes a parabola that opens to the left.

- **Vertex:** The vertex \((h, k)\) is given by \((-0.25, -2)\).
- **Focus:** The standard form gives \( a = \frac{1}{16} \). The focus is at \((h - a, k)\), so \((-0.25 - \frac{1}{16}, -2) = (-0.3125, -2)\).
- **Axis of Symmetry:** The axis of symmetry is the horizontal line \( y = -2 \).
- **Directrix:** The directrix is the vertical line \( x = h + a \), so \( x = -0.25 + \frac{1}{16} = -0.1875 \).

### 3. \( -(1/3)(x-3) = (y + 5)^2 \)

This equation can be rewritten to match the standard form \( (y-k)^2 = -4a(x-h) \).

- **Vertex:** The vertex \((h, k)\) is given by \((3, -5)\).
- **Focus:** From the standard form, \( -\frac{1}{3} = -4a \), so \( a = \frac{1}{12} \). The focus is at \((h - a, k)\), so \((3 - \frac{1}{12}, -5) = \left(\frac{35}{12}, -5\right)\).
- **Axis of Symmetry:** The axis of symmetry is the horizontal line \( y = -5 \).
- **Directrix:** The directrix is the vertical line \( x = h + a \), so \( x = 3 + \frac{1}{12} = \frac{37}{12} \).

### 4. \( -(y + 2) = (x - 2)^2 \)

This equation matches the standard form \( (x-h)^2 = -4a(y-k) \), which describes a parabola that opens downward.

- **Vertex:** The vertex \((h, k)\) is given by \((2, -2)\).
- **Focus:** Here, \( -1 = -4a \), so \( a = \frac{1}{4} \). The focus is at \((h, k - a)\), so \((2, -2 - \frac{1}{4}) = (2, -2.25)\).
- **Axis of Symmetry:** The axis of symmetry is the vertical line \( x = 2 \).
- **Directrix:** The directrix is the line \( y = k + a \), so \( y = -2 + \frac{1}{4} = -1.75 \).

### 5. \( 3y + 4x = -(2x^2) - 14 \)

This equation needs to be rewritten to fit a standard form. Start by rewriting it as:
\[ 2x^2 + 4x + 3y = -14 \]
\[ 3y = -2x^2 - 4x - 14 \]
\[ y = -\frac{2}{3}x^2 - \frac{4}{3}x - \frac{14}{3} \]

This is a parabola in the form \( y = ax^2 + bx + c \). Let's complete the square to find the vertex form:
\[ y = -\frac{2}{3}\left(x^2 + 2x + 1\right) - \frac{14}{3} + \frac{2}{3} = -\frac{2}{3}(x+1)^2 - \frac{12}{3} \]

Now, rewrite as:
\[ y = -\frac{2}{3}(x+1)^2 - 4 \]

The standard form is \( (x-h)^2 = -4a(y-k) \).

- **Vertex:** The vertex \((h, k)\) is \((-1, -4)\).
- **Focus:** Here, \( -\frac{2}{3} = -4a \), so \( a = \frac{1}{6} \). The focus is at \((-1, -4 + \frac{1}{6}) = (-1, -3.83)\).
- **Axis of Symmetry:** The axis of symmetry is the vertical line \( x = -1 \).
- **Directrix:** The directrix is the line \( y = -4 - \frac{1}{6} = -4.17 \).

### 6. \( x = -2(y + 2)^2 \)

This equation matches the standard form \( (y-k)^2 = -4a(x-h) \), describing a parabola that opens to the left.

- **Vertex:** The vertex \((h, k)\) is \((-2, -2)\).
- **Focus:** Here, \( a = \frac{1}{8} \). The focus is at \((h - a, k)\), so \((-2 - \frac{1}{8}, -2) = (-2.125, -2)\).
- **Axis of Symmetry:** The axis of symmetry is the horizontal line \( y = -2 \).
- **Directrix:** The directrix is the vertical line \( x = h + a \), so \( x = -2 + \frac{1}{8} = -1.875 \).

### Graph Sketches

For each parabola:
- Start by plotting the vertex.
- Sketch the parabola opening in the correct direction (up, down, left, or right).
- Mark the focus and draw the directrix, ensuring the focus lies inside the parabola and the directrix lies outside, opposite the vertex.

**Do you need help with sketching these graphs or calculating additional details?**

### Next Steps:
1. Would you like to discuss the effects of the parameter \( a \) on the shape of each parabola?
2. Do you need a step-by-step guide on completing the square?
3. Should we analyze how these parabolas behave in specific applications?
4. Would you like more examples or different types of conic sections?
5. Do you want to explore the intersections of these parabolas with other curves?

**Tip:** When identifying the focus and directrix, remember they are always equidistant from any point on the parabola.

Learn how to identify vertex, focus, axis of symmetry, and directrix of parabolas in standard forms (x-h)^2 = 4a(y-k), (y-k)^2 = -4a(x-h). Detailed solutions and graph sketches provided.

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