Graph the quadratic function

f left parenthesis x right parenthesis equals negative 2 x squared plus 8 x minus 12f(x)=−2x2+8x−12.

Give the ​(a)​ vertex, ​(b)​ axis, ​(c)​ domain, and ​(d) range.

Question content area bottom left

Part 1

​(a) The vertex is

enter your response here.

​(Type an ordered​ pair.)

Part 2

Use the graphing tool to graph the function.

Click to
enlarge
graph

Part 3

​(b) The axis is

enter your response here.

​(Type an​ equation.)

Part 4

​(c) The domain is

enter your response here.

​(Type your answer in interval​ notation.)

Part 5

​(d) The range is

enter your response here.

​(Type your answer in interval​ notation.)

To analyze the quadratic function \( f(x) = -2x^2 + 8x - 12 \), let's go through each part of the problem.

### Part 1: Finding the Vertex
The vertex form of a quadratic function \( f(x) = ax^2 + bx + c \) has its vertex at the point:
\[
\left( x, y \right) = \left( \frac{-b}{2a}, f\left( \frac{-b}{2a} \right) \right)
\]

For \( f(x) = -2x^2 + 8x - 12 \), we have:
- \( a = -2 \)
- \( b = 8 \)

1. Calculate \( x \)-coordinate of the vertex:
   \[
   x = \frac{-b}{2a} = \frac{-8}{2 \cdot -2} = \frac{-8}{-4} = 2
   \]

2. Substitute \( x = 2 \) into \( f(x) \) to find the \( y \)-coordinate of the vertex:
   \[
   f(2) = -2(2)^2 + 8(2) - 12 = -2 \cdot 4 + 16 - 12 = -8 + 16 - 12 = -4
   \]

Thus, the vertex is at:
\[
\text{Vertex} = (2, -4)
\]

### Part 2: Graphing the Function
The graph of this quadratic function will be a downward-opening parabola, as the leading coefficient \( a = -2 \) is negative. 

### Part 3: Axis of Symmetry
The axis of symmetry for a quadratic function is the vertical line passing through the vertex. So the axis of symmetry here is:
\[
x = 2
\]

### Part 4: Domain
For any quadratic function, the domain is all real numbers because \( x \) can take any value. Thus:
\[
\text{Domain} = (-\infty, \infty)
\]

### Part 5: Range
Since the parabola opens downward, the function has a maximum value at the vertex. The maximum \( y \)-value is the \( y \)-coordinate of the vertex, which is \( -4 \). Thus:
\[
\text{Range} = (-\infty, -4]
\]

### Summary of Answers
- **(a) Vertex:** \( (2, -4) \)
- **(b) Axis:** \( x = 2 \)
- **(c) Domain:** \( (-\infty, \infty) \)
- **(d) Range:** \( (-\infty, -4] \)

Would you like a detailed graph of this function?

#### Relative Questions
1. How do you determine the direction a parabola opens based on the coefficient \( a \)?
2. What is the significance of the vertex in a quadratic function?
3. How does the value of \( b \) affect the location of the axis of symmetry?
4. How would the range change if the parabola opened upwards?
5. Can you have a different domain for a quadratic function under any circumstances?

#### Tip
Remember, for any quadratic function, the axis of symmetry will always be \( x = \frac{-b}{2a} \), dividing the parabola into two symmetrical halves.

Graph the quadratic function f(x) = −2x² + 8x − 12. Give the (a) vertex, (b) axis, (c) domain, and (d) range.

Explore how to graph and analyze the quadratic function f(x) = −2x² + 8x − 12. This solution covers finding the vertex, axis of symmetry, domain, and range, and demonstrates key algebraic concepts suitable for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation