Determine the focus and directrix of the parabola with the given equation. Sketch the graph, and indicate the focus, directrix, and vertex if the equation is 5. x^2- 6x - 2y + 9 = 0

To determine the focus, directrix, and vertex of the given parabola, we need to rewrite the equation in its standard form. Let's go through the steps:

### Given Equation:
\[ x^2 - 6x - 2y + 9 = 0 \]

### Step 1: Rearrange the equation to isolate the quadratic terms.
\[ x^2 - 6x = 2y - 9 \]

### Step 2: Complete the square on the \(x\)-terms.
To complete the square, we add and subtract \(\left(\frac{-6}{2}\right)^2 = 9\) on the left side:
\[ x^2 - 6x + 9 = 2y - 9 + 9 \]

This simplifies to:
\[ (x - 3)^2 = 2y \]

### Step 3: Rewrite the equation in standard form.
The equation is now:
\[ (x - 3)^2 = 2(y - 0) \]

This is in the standard form of a vertical parabola:
\[ (x - h)^2 = 4p(y - k) \]

Where \((h, k)\) is the vertex, and \(p\) is the distance from the vertex to the focus (or directrix).

### Step 4: Identify the vertex, focus, and directrix.
- **Vertex (h, k):** The vertex is \((3, 0)\).
- **Value of \(p\):** From \(4p = 2\), we get \(p = \frac{1}{2}\).
- **Focus:** Since the parabola opens upwards (positive \(p\)), the focus is \(\frac{1}{2}\) units above the vertex. The focus is at \((3, \frac{1}{2})\).
- **Directrix:** The directrix is a horizontal line \(\frac{1}{2}\) units below the vertex, which is \(y = -\frac{1}{2}\).

### Step 5: Sketch the Graph
To sketch the graph:
1. Plot the vertex at \((3, 0)\).
2. Plot the focus at \((3, \frac{1}{2})\).
3. Draw the directrix as the line \(y = -\frac{1}{2}\).
4. Sketch the parabola opening upwards.

I can generate a visual representation of this parabola for you if you'd like. Would you like me to do that?

### Relative Questions:
1. How does completing the square help in rewriting quadratic equations?
2. Why is the vertex of a parabola important in determining its graph?
3. How does the sign of \(p\) affect the direction in which the parabola opens?
4. What is the relationship between the focus and directrix in a parabola?
5. How can you derive the equation of a parabola given its focus and directrix?

### Tip:
Remember that the distance from the vertex to the focus and to the directrix is always the same, which is \( |p| \) in the equation.

Learn how to determine the focus, directrix, and vertex of a parabola with the equation x^2 - 6x - 2y + 9 = 0. This problem involves completing the square to rewrite the equation in standard form and identifying key geometric properties of the parabola. Suitable for students in Grades 9-12.

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