A quadratic function f is given.
f(x) = −x2 + 16x
(a)
Express f in vertex form.
f(x) =
Incorrect: Your answer is incorrect.
(b)
Find the vertex and x- and y-intercepts of f. (If an answer does not exist, enter DNE.)
vertex
(x, y)
=

x-intercept (smaller x-value)
(x, y)
=

x-intercept (larger x-value)
(x, y)
=

y-intercept
(x, y)
=

(c)
Sketch a graph of f.
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Segment
Circle
Vertical Parabola
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(d)
Find the domain and range of f. (Enter your answers using interval notation.)
domain
range
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Let's go step by step to solve the given quadratic function problem. The quadratic function provided is:

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

### (a) Express \( f(x) \) in vertex form

The vertex form of a quadratic function is:

\[
f(x) = a(x - h)^2 + k
\]

Where \( (h, k) \) is the vertex of the parabola. To convert the given equation to vertex form, we will complete the square.

1. **Factor out the coefficient of \( x^2 \) from the quadratic and linear terms:**

\[
f(x) = -(x^2 - 16x)
\]

2. **Complete the square:**
   - Take the coefficient of \( x \), divide it by 2, and square it. In this case:
   \[
   \left(\frac{-16}{2}\right)^2 = (-8)^2 = 64
   \]
   - Add and subtract 64 inside the parentheses:

\[
f(x) = -(x^2 - 16x + 64 - 64)
\]
\[
f(x) = -( (x - 8)^2 - 64)
\]

3. **Simplify:**
   - Distribute the negative sign:

\[
f(x) = -(x - 8)^2 + 64
\]

Thus, the vertex form of the function is:

\[
f(x) = -(x - 8)^2 + 64
\]

### (b) Find the vertex and intercepts

#### Vertex:
From the vertex form \( f(x) = -(x - 8)^2 + 64 \), the vertex \( (h, k) \) is \( (8, 64) \).

**Vertex:**
\[
(x, y) = (8, 64)
\]

#### X-intercepts:
To find the x-intercepts, set \( f(x) = 0 \) and solve for \( x \):

\[
-x^2 + 16x = 0
\]
Factor the equation:

\[
-x(x - 16) = 0
\]

So the x-intercepts are \( x = 0 \) and \( x = 16 \).

**X-intercepts:**
\[
(0, 0), \quad (16, 0)
\]

#### Y-intercept:
The y-intercept occurs when \( x = 0 \). Substituting \( x = 0 \) into the original equation:

\[
f(0) = -(0)^2 + 16(0) = 0
\]

**Y-intercept:**
\[
(0, 0)
\]

### (c) Sketch a graph of \( f(x) \)
- The parabola opens **downward** because the coefficient of \( x^2 \) is negative.
- The vertex is \( (8, 64) \).
- The x-intercepts are at \( (0, 0) \) and \( (16, 0) \).
- The y-intercept is at \( (0, 0) \).

The parabola will be symmetric about \( x = 8 \), and the graph should be a downward-facing parabola with the vertex at \( (8, 64) \).

### (d) Domain and range

- The **domain** of any quadratic function is all real numbers:
\[
\text{Domain: } (-\infty, \infty)
\]
- The **range** is determined by the vertex and the direction the parabola opens. Since the parabola opens downward and the maximum value is 64 (at the vertex), the range is:
\[
\text{Range: } (-\infty, 64]
\]

### Final Answers:
- **(a)** Vertex form: \( f(x) = -(x - 8)^2 + 64 \)
- **(b)** 
  - Vertex: \( (8, 64) \)
  - X-intercepts: \( (0, 0) \) and \( (16, 0) \)
  - Y-intercept: \( (0, 0) \)
- **(d)** 
  - Domain: \( (-\infty, \infty) \)
  - Range: \( (-\infty, 64] \)

Would you like more details on any of the steps? Here are some related questions to explore further:

1. How can completing the square help in finding the vertex form of a quadratic function?
2. What is the significance of the vertex in the graph of a quadratic function?
3. How do we determine if a parabola opens upward or downward?
4. Why does every quadratic function have a domain of \( (-\infty, \infty) \)?
5. Can a quadratic function have more than two x-intercepts? Why or why not?

**Tip:** To check if a quadratic opens upward or downward, just look at the sign of the coefficient of \( x^2 \). If it's positive, the parabola opens upward; if it's negative, it opens downward.

A quadratic function f is given.
f(x) = −x^2 + 16x
(a) Express f in vertex form.
(b) Find the vertex and x- and y-intercepts of f. (If an answer does not exist, enter DNE.)
(c) Sketch a graph of f.
(d) Find the domain and range of f. (Enter your answers using interval notation.)

This page explains how to express the quadratic function f(x) = -x² + 16x in vertex form, find the vertex, x- and y-intercepts, and determine the domain and range. It includes step-by-step solutions suitable for high school students (Grades 9-11) studying quadratic functions.

Math Problem Statement

A quadratic function f is given. f(x) = −x2 + 16x (a) Express f in vertex form. f(x) = Incorrect: Your answer is incorrect. (b) Find the vertex and x- and y-intercepts of f. (If an answer does not exist, enter DNE.) vertex (x, y)

x-intercept (smaller x-value) (x, y)

x-intercept (larger x-value) (x, y)

y-intercept (x, y)

Solution

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