Trigonometry (MindTap Course List)
10th Edition
ISBN:9781337278461
Author:Ron Larson
Publisher:Ron Larson
ChapterP: Prerequisites
SectionP.7: A Library Of Parent Functions
Problem 47E
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In need of help with b and c
![**Problem 3: Given \( f(x) = -x^2 + 4x + 12 \), \( 0 \leq x \leq 8 \):**
**(a) Carefully, graph \( f(x) \), \( 0 \leq x \leq 8 \).**
**Solution:**
To graph the function \( f(x) = -x^2 + 4x + 12 \) for the domain \( 0 \leq x \leq 8 \), follow these steps:
1. **Identify Key Points:**
- **Vertex:** The vertex of a quadratic function \( ax^2 + bx + c \) can be found using \( x = -\frac{b}{2a} \).
For \( f(x) = -x^2 + 4x + 12 \):
\[
a = -1, \quad b = 4, \quad c = 12
\]
\[
x = -\frac{4}{2(-1)} = 2
\]
Substituting \( x = 2 \) into \( f(x) \) to find \( y \):
\[
f(2) = -(2)^2 + 4(2) + 12 = -4 + 8 + 12 = 16
\]
The vertex is at \( (2, 16) \).
2. **Find the Y-Intercept:**
To find the y-intercept, set \( x = 0 \):
\[
f(0) = -(0)^2 + 4(0) + 12 = 12
\]
So, the y-intercept is \( (0, 12) \).
3. **Find the X-Intercepts:**
To find the x-intercepts, set \( f(x) = 0 \):
\[
-x^2 + 4x + 12 = 0
\]
Solving the quadratic equation:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
\[
x = \frac{-4 \pm \sqrt{16 + 48}}{-2}
\]
\[
x = \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9d19ae5-a74d-4667-9411-09fe8f4f758b%2Fa7c5cfcc-c419-4f2e-bb36-01432f8074ee%2Fx0nvz3.png&w=3840&q=75)
Transcribed Image Text:**Problem 3: Given \( f(x) = -x^2 + 4x + 12 \), \( 0 \leq x \leq 8 \):**
**(a) Carefully, graph \( f(x) \), \( 0 \leq x \leq 8 \).**
**Solution:**
To graph the function \( f(x) = -x^2 + 4x + 12 \) for the domain \( 0 \leq x \leq 8 \), follow these steps:
1. **Identify Key Points:**
- **Vertex:** The vertex of a quadratic function \( ax^2 + bx + c \) can be found using \( x = -\frac{b}{2a} \).
For \( f(x) = -x^2 + 4x + 12 \):
\[
a = -1, \quad b = 4, \quad c = 12
\]
\[
x = -\frac{4}{2(-1)} = 2
\]
Substituting \( x = 2 \) into \( f(x) \) to find \( y \):
\[
f(2) = -(2)^2 + 4(2) + 12 = -4 + 8 + 12 = 16
\]
The vertex is at \( (2, 16) \).
2. **Find the Y-Intercept:**
To find the y-intercept, set \( x = 0 \):
\[
f(0) = -(0)^2 + 4(0) + 12 = 12
\]
So, the y-intercept is \( (0, 12) \).
3. **Find the X-Intercepts:**
To find the x-intercepts, set \( f(x) = 0 \):
\[
-x^2 + 4x + 12 = 0
\]
Solving the quadratic equation:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
\[
x = \frac{-4 \pm \sqrt{16 + 48}}{-2}
\]
\[
x = \frac
![### Calculus Problem: Finding Net Area and Evaluating Integrals
#### Problem Statement
(b) Find the net area from \( x = 0 \) and \( x = 8 \).
(c) Evaluate \( \int_{0}^{8} \lvert f(x) \rvert \, dx \).
#### Solution Approach
To solve these problems, we need to understand the concept of net area and the integral of the absolute value of a function over an interval.
**Net Area:**
The net area between a function \( f(x) \) and the x-axis from \( x = 0 \) to \( x = 8 \) is found by evaluating the definite integral:
\[ \text{Net Area} = \int_{0}^{8} f(x) \, dx \]
The net area considers the function's regions above the x-axis as positive and regions below the x-axis as negative.
**Area under the Absolute Value Function:**
To find the total area between the function \( f(x) \) and the x-axis considering all regions as positive (regardless of the function dipping below the x-axis), we evaluate:
\[ \int_{0}^{8} \lvert f(x) \rvert \, dx \]
This integral sums the absolute values of the function over the interval, treating all areas as positive.
### Summary
- The net area from \( x = 0 \) to \( x = 8 \) is the integral \( \int_{0}^{8} f(x) \, dx \).
- The total area, treating all regions as positive, is \( \int_{0}^{8} \lvert f(x) \rvert \, dx \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9d19ae5-a74d-4667-9411-09fe8f4f758b%2Fa7c5cfcc-c419-4f2e-bb36-01432f8074ee%2Fi4bfibl.png&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Finding Net Area and Evaluating Integrals
#### Problem Statement
(b) Find the net area from \( x = 0 \) and \( x = 8 \).
(c) Evaluate \( \int_{0}^{8} \lvert f(x) \rvert \, dx \).
#### Solution Approach
To solve these problems, we need to understand the concept of net area and the integral of the absolute value of a function over an interval.
**Net Area:**
The net area between a function \( f(x) \) and the x-axis from \( x = 0 \) to \( x = 8 \) is found by evaluating the definite integral:
\[ \text{Net Area} = \int_{0}^{8} f(x) \, dx \]
The net area considers the function's regions above the x-axis as positive and regions below the x-axis as negative.
**Area under the Absolute Value Function:**
To find the total area between the function \( f(x) \) and the x-axis considering all regions as positive (regardless of the function dipping below the x-axis), we evaluate:
\[ \int_{0}^{8} \lvert f(x) \rvert \, dx \]
This integral sums the absolute values of the function over the interval, treating all areas as positive.
### Summary
- The net area from \( x = 0 \) to \( x = 8 \) is the integral \( \int_{0}^{8} f(x) \, dx \).
- The total area, treating all regions as positive, is \( \int_{0}^{8} \lvert f(x) \rvert \, dx \).
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