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:

Given:

$f(x) = -x^2 + 6x - 9$

This is a quadratic function in standard form $f(x) = ax^2 + bx + c$, where:

• $a = -1$
• $b = 6$
• $c = -9$

1. Finding the Vertex:

The vertex of a parabola in standard form can be found using the formula: $x = \frac{-b}{2a}$

Substitute the values of $a$ and $b$: $x = \frac{-6}{2(-1)} = \frac{-6}{-2} = 3$

Now substitute $x = 3$ back into the original equation to find the $y$-coordinate of the vertex: $f(3) = -(3)^2 + 6(3) - 9$ $f(3) = -9 + 18 - 9 = 0$

Thus, the vertex is at $(3, 0)$.

2. Finding the y-intercept:

The $y$-intercept occurs when $x = 0$. Substitute $x = 0$ into the function: $f(0) = -(0)^2 + 6(0) - 9 = -9$

Thus, the $y$-intercept is $y = -9$.

3. Finding the x-intercepts:

The $x$-intercepts occur when $f(x) = 0$. So, set the equation equal to 0 and solve for $x$: $-x^2 + 6x - 9 = 0$

This is a quadratic equation, which can be solved using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substitute $a = -1$, $b = 6$, and $c = -9$: $x = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-9)}}{2(-1)}$ $x = \frac{-6 \pm \sqrt{36 - 36}}{-2}$ $x = \frac{-6 \pm \sqrt{0}}{-2}$ $x = \frac{-6 \pm 0}{-2}$ $x = 3$

Since there is only one solution, there is one x-intercept at $x = 3$.

Final Results:

• Vertex: $(3, 0)$
• y-intercept: $y = -9$
• x-intercept: $x = 3$

The graph is a downward-facing parabola with its vertex at $(3, 0)$.

Would you like further details on any part of the solution?

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Relative Questions:

1. How do we derive the quadratic formula for solving quadratic equations?
2. What is the significance of the vertex in the graph of a parabola?
3. Can a quadratic equation have more than two x-intercepts? Why or why not?
4. What changes in the graph if the coefficient $a$ is positive?
5. How can we determine whether the vertex represents a maximum or minimum point?

Tip:

When the coefficient of $x^2$ (i.e., $a$) is negative, the parabola always opens downward, making the vertex the maximum point of the graph.