Sketch the graph of the quadratic function. f(x) = −x2 + 6x − 9 Selection Tool Line Ray Segment Circle Vertical Parabola Horizontal Parabola Point No Solution Help -10-9-8-7-6-5-4-3-2-112345678910-10-9-8-7-6-5-4-3-2-112345678910 Clear Graph Delete Layer Fill WebAssign Graphing Tool Graph LayersToggle Open/Closed After you add an object to the graph you can use Graph Layers to view and edit its properties. 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)

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?

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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.