Sketch the graph of the quadratic function.
f(x) = −x2 + 6x − 9
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Indicate the coordinates of the vertex, the y-intercept and the x-intercepts (if any). (If an answer does not exist, enter DNE. If there are multiple x-intercepts, enter your answers as a comma-separated list.)
vertex	(x, y)	
= 

  

y-intercept	y	=	
x-intercept	x	=	
 

Let's start by analyzing the quadratic function \( f(x) = -x^2 + 6x - 9 \) and sketch its graph, including the vertex, y-intercept, and x-intercepts.

### 1. **Vertex of the Parabola**

The equation of a quadratic function is generally given by:
\[
f(x) = ax^2 + bx + c
\]
In this case:
- \( a = -1 \)
- \( b = 6 \)
- \( c = -9 \)

The vertex of a parabola given by \( f(x) = ax^2 + bx + c \) occurs at:
\[
x_{\text{vertex}} = \frac{-b}{2a}
\]
Substitute \( a = -1 \) and \( b = 6 \):
\[
x_{\text{vertex}} = \frac{-6}{2(-1)} = \frac{-6}{-2} = 3
\]

Now, find the \( y \)-coordinate of the vertex by substituting \( x = 3 \) into the function:
\[
f(3) = -(3)^2 + 6(3) - 9 = -9 + 18 - 9 = 0
\]
Thus, the vertex is \( (3, 0) \).

### 2. **Y-intercept**

The y-intercept occurs where \( x = 0 \). Substituting \( x = 0 \) into the function:
\[
f(0) = -(0)^2 + 6(0) - 9 = -9
\]
Thus, the y-intercept is \( y = -9 \).

### 3. **X-intercepts**

The x-intercepts occur where \( f(x) = 0 \). Set the function equal to zero:
\[
-x^2 + 6x - 9 = 0
\]
This is a quadratic equation. Solve it using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substitute \( a = -1 \), \( b = 6 \), and \( c = -9 \):
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-9)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 36}}{-2}
\]
\[
x = \frac{-6 \pm \sqrt{0}}{-2} = \frac{-6}{-2} = 3
\]
Thus, the only x-intercept is \( x = 3 \).

### Summary:
- **Vertex**: \( (3, 0) \)
- **Y-intercept**: \( y = -9 \)
- **X-intercept**: \( x = 3 \) (only one intercept)

### Graph of the function:
- The parabola opens **downward** because \( a = -1 \), and the vertex is the highest point on the graph.
- The vertex is at \( (3, 0) \), and it touches the x-axis at that point.
- The y-intercept is at \( (0, -9) \).

If you'd like me to generate a graph, let me know! Would you like a detailed sketch?

---

Here are 5 questions that might help expand on this concept:
1. What is the significance of the vertex in the graph of a quadratic function?
2. How does the value of \( a \) affect the direction in which the parabola opens?
3. How would the graph change if \( a \) were positive?
4. Can a quadratic function have more than two x-intercepts? Why or why not?
5. How does completing the square help in finding the vertex?

**Tip**: The discriminant \( \Delta = b^2 - 4ac \) in the quadratic formula helps determine the number of x-intercepts (real roots). If \( \Delta = 0 \), there is exactly one x-intercept.

Sketch the graph of the quadratic function.
f(x) = -x^2 + 6x - 9
Indicate the coordinates of the vertex, the y-intercept and the x-intercepts (if any). (If an answer does not exist, enter DNE. If there are multiple x-intercepts, enter your answers as a comma-separated list.)

Learn how to sketch the graph of the quadratic function f(x) = -x^2 + 6x - 9. This problem covers key algebraic concepts, including finding the vertex, y-intercept, and x-intercept using the quadratic formula and vertex formula. Suitable for students in Grades 9-11.

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