Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![### Question 6
Given the parabola \( y = -2x^2 - 8x + 9 \). State the Range. *(Draw a sketch if necessary)*
#### Explanation:
To determine the range of a quadratic function, it's helpful to identify the vertex of the parabola, especially since the function is written in standard form \( y= ax^2 + bx + c \).
1. **Identify the coefficients:** In the given function \( y = -2x^2 - 8x + 9 \):
- \( a = -2 \)
- \( b = -8 \)
- \( c = 9 \)
2. **Calculate the vertex:** The x-coordinate of the vertex of a parabola in standard form can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values of \( a \) and \( b \):
\[
x = -\frac{-8}{2(-2)} = \frac{8}{-4} = -2
\]
3. **Find the y-coordinate of the vertex:** Substitute \( x = -2 \) back into the original equation to find the y-coordinate.
\[
y = -2(-2)^2 - 8(-2) + 9 = -2(4) + 16 + 9 = -8 + 16 + 9 = 17
\]
The vertex of the parabola is at \( (-2, 17) \). Since the coefficient of \( x^2 \) is negative (\( a = -2 \)), the parabola opens downwards, hence the vertex represents the maximum point.
4. **State the range:**
- The range includes all values of \( y \) that are less than or equal to the maximum value (the y-coordinate of the vertex).
\[
\text{Range} = (-\infty, 17]
\]
Drawing a sketch would show a downward-opening parabola with its highest point at (–2, 17).
#### Graph/Diagram:
If necessary, a simple sketch of the parabola would display the following:
- A parabolic curve that opens downwards.
- The vertex at (-2, 17), indicating the highest point on the graph.
- The y-axis and x-axis to reference the position](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F661c5eba-7c52-43bd-a2b1-7e7eb4bf2684%2Feeb2045e-7651-476a-9a3a-22995251eda1%2F128wn29_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 6
Given the parabola \( y = -2x^2 - 8x + 9 \). State the Range. *(Draw a sketch if necessary)*
#### Explanation:
To determine the range of a quadratic function, it's helpful to identify the vertex of the parabola, especially since the function is written in standard form \( y= ax^2 + bx + c \).
1. **Identify the coefficients:** In the given function \( y = -2x^2 - 8x + 9 \):
- \( a = -2 \)
- \( b = -8 \)
- \( c = 9 \)
2. **Calculate the vertex:** The x-coordinate of the vertex of a parabola in standard form can be found using the formula \( x = -\frac{b}{2a} \). Substituting the values of \( a \) and \( b \):
\[
x = -\frac{-8}{2(-2)} = \frac{8}{-4} = -2
\]
3. **Find the y-coordinate of the vertex:** Substitute \( x = -2 \) back into the original equation to find the y-coordinate.
\[
y = -2(-2)^2 - 8(-2) + 9 = -2(4) + 16 + 9 = -8 + 16 + 9 = 17
\]
The vertex of the parabola is at \( (-2, 17) \). Since the coefficient of \( x^2 \) is negative (\( a = -2 \)), the parabola opens downwards, hence the vertex represents the maximum point.
4. **State the range:**
- The range includes all values of \( y \) that are less than or equal to the maximum value (the y-coordinate of the vertex).
\[
\text{Range} = (-\infty, 17]
\]
Drawing a sketch would show a downward-opening parabola with its highest point at (–2, 17).
#### Graph/Diagram:
If necessary, a simple sketch of the parabola would display the following:
- A parabolic curve that opens downwards.
- The vertex at (-2, 17), indicating the highest point on the graph.
- The y-axis and x-axis to reference the position
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