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General form and find the vertex, focus, LR1, LR2, Directrix, Axis of symmetry, length of LR and graph

The equation given in the image is:

\[
x^2 + 16y + 4x + 36 = 0
\]

### Step 1: Rewrite the equation in standard form (completing the square).
We start by rearranging the terms to group the \(x\) terms and \(y\) terms:

\[
x^2 + 4x + 16y + 36 = 0
\]

### Step 2: Completing the square for the \(x\) terms.
Take the \(x\) terms: \(x^2 + 4x\). To complete the square, we follow these steps:
1. Take half of the coefficient of \(x\), which is \(\frac{4}{2} = 2\), and square it: \(2^2 = 4\).
2. Add and subtract 4 inside the equation:

\[
(x^2 + 4x + 4) + 16y + 36 - 4 = 0
\]

This simplifies to:

\[
(x + 2)^2 + 16y + 32 = 0
\]

### Step 3: Solve for \(y\) to get the equation in standard form.
Now subtract 32 from both sides:

\[
(x + 2)^2 + 16y = -32
\]

Now, divide the equation by 16 to isolate \(y\):

\[
y = \frac{-(x + 2)^2 - 32}{16}
\]

Simplifying:

\[
y = -\frac{1}{16}(x + 2)^2 - 2
\]

This is the equation in standard form of a parabola.

### Step 4: Identifying key elements.

1. **Vertex**: From the standard form \(y = -\frac{1}{16}(x + 2)^2 - 2\), the vertex is \((-2, -2)\).
2. **Axis of symmetry**: The axis of symmetry is a vertical line that passes through the vertex, so \(x = -2\).
3. **Focus**: The general form for a parabola is \(y = a(x - h)^2 + k\), where \(a = -\frac{1}{16}\). The distance from the vertex to the focus is \(\frac{1}{4a}\), so the focus is at:
   \[
   \frac{1}{4(-\frac{1}{16})} = -4
   \]
   The focus is 4 units below the vertex, so the focus is at \((-2, -6)\).
4. **Directrix**: The directrix is a line that is the same distance from the vertex as the focus but in the opposite direction. Therefore, the directrix is 4 units above the vertex at \(y = 2\).
5. **Length of the Latus Rectum (LR)**: The length of the latus rectum is \(\frac{1}{|a|} = 16\).

Would you like more details or clarifications? Here are five related questions to consider:

1. What are the properties of parabolas with horizontal axes of symmetry?
2. How does changing the coefficient of \(x^2\) affect the width of the parabola?
3. How can you verify the location of the directrix graphically?
4. Why is the length of the latus rectum connected to the parameter \(a\)?
5. How do you graph a parabola step-by-step from its standard form?

**Tip**: When completing the square, always remember to balance the equation by adding and subtracting the same value!

Find the general form, vertex, focus, latus rectum, directrix, axis of symmetry, length of the LR, and graph of the equation x^2 + 16y + 4x + 36 = 0.

Explore the parabola equation x^2 + 16y + 4x + 36 = 0 by finding its vertex, focus, directrix, axis of symmetry, and the length of the latus rectum. Learn how to rewrite the equation in standard form and use key formulas to solve for these characteristics.

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