Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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please don’t forget any steps! thank you!
![**Lesson: Solving Quadratic Equations Using Given Points**
**Objective:** Learn how to find a quadratic equation given three points through which its graph passes.
**Problem Statement:**
- (15 points) Find the quadratic equation \( y = ax^2 + bx + c \) whose graph passes through the points: \((-3, 1)\), \((-1, -7)\), \((1, 1)\).
**Solution Steps:**
1. **Substitute the Points into the Equation:**
Each point \((x, y)\) is substituted into the quadratic equation \( y = ax^2 + bx + c \).
- For \((-3, 1)\):
\[
a(-3)^2 + b(-3) + c = 1
\]
Simplified:
\[
9a - 3b + c = 1
\]
- For \((-1, -7)\):
\[
a(-1)^2 + b(-1) + c = -7
\]
Simplified:
\[
a - b + c = -7
\]
- For \((1, 1)\):
\[
a(1)^2 + b(1) + c = 1
\]
Simplified:
\[
a + b + c = 1
\]
2. **Set Up the System of Equations:**
- \(9a - 3b + c = 1\)
- \(a - b + c = -7\)
- \(a + b + c = 1\)
3. **Solve the System of Equations:**
To solve for \(a\), \(b\), and \(c\), you can use methods such as substitution or elimination to find the values that satisfy all equations simultaneously.
Graphical representation and further solution steps are often used to ensure understanding of the procedures for finding such quadratic equations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64a9bba9-6d54-4904-907b-a62783e46f3d%2Ff24ebe1e-a58d-43e2-b05e-6bfe761a27cf%2F58nm66_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Lesson: Solving Quadratic Equations Using Given Points**
**Objective:** Learn how to find a quadratic equation given three points through which its graph passes.
**Problem Statement:**
- (15 points) Find the quadratic equation \( y = ax^2 + bx + c \) whose graph passes through the points: \((-3, 1)\), \((-1, -7)\), \((1, 1)\).
**Solution Steps:**
1. **Substitute the Points into the Equation:**
Each point \((x, y)\) is substituted into the quadratic equation \( y = ax^2 + bx + c \).
- For \((-3, 1)\):
\[
a(-3)^2 + b(-3) + c = 1
\]
Simplified:
\[
9a - 3b + c = 1
\]
- For \((-1, -7)\):
\[
a(-1)^2 + b(-1) + c = -7
\]
Simplified:
\[
a - b + c = -7
\]
- For \((1, 1)\):
\[
a(1)^2 + b(1) + c = 1
\]
Simplified:
\[
a + b + c = 1
\]
2. **Set Up the System of Equations:**
- \(9a - 3b + c = 1\)
- \(a - b + c = -7\)
- \(a + b + c = 1\)
3. **Solve the System of Equations:**
To solve for \(a\), \(b\), and \(c\), you can use methods such as substitution or elimination to find the values that satisfy all equations simultaneously.
Graphical representation and further solution steps are often used to ensure understanding of the procedures for finding such quadratic equations.
Transcribed Image Text:**Graphing Systems of Inequalities**
The image presents a system of inequalities for graphing:
1. \( 21x + 7y > 14 \)
2. \( 4x^2 - 4y > -8 \)
**Graph Explanation:**
The graph area appears to be part of a coordinate plane focused on the \( y \)-axis, indicating vertical values ranging from 8 to 15. It includes grid lines to assist in plotting points accurately.
**How to Approach Graphing:**
1. **Inequality 1: \( 21x + 7y > 14 \)**
- Rearrange it to slope-intercept form \( y > -\frac{3}{1}x + 2 \).
- Plot the line \( y = -3x + 2 \). Since the inequality is '>', shade above the line.
2. **Inequality 2: \( 4x^2 - 4y > -8 \)**
- Simplify it to \( y < x^2 + 2 \).
- Plot the curve \( y = x^2 + 2 \). Since the inequality is '<', shade below the curve.
**Graphing Strategy:**
- Use dashed lines to show boundaries where the values of \( y \) do not include the boundary lines themselves (as the inequalities are strict inequalities '<>' rather than '≤' or '≥').
- Shade the appropriate regions as per inequalities.
- The feasible region is where the shaded areas overlap.
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