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
![**Quadratic Conversion and Analysis**
**Objective**: Convert the quadratic into standard form by detailing the steps involved in changing forms. Then, identify the roots and vertex of the given quadratic equation.
**Problem Statement**:
Given the quadratic equation:
\[ y = 3x(x - 2) \]
**Instructions**:
1. **Expand the Equation**: Distribute the terms to convert the equation into standard form.
2. **Standard Form**: Express the quadratic in the standard form of \( ax^2 + bx + c \).
3. **Identify Roots**: Solve for the values of \( x \) where \( y = 0 \).
4. **Find the Vertex**: Calculate the vertex of the parabola using the standard form.
**Steps to Solution**:
- *Expanding the Quadratic*:
\[ y = 3x^2 - 6x \]
- *Standard Form*:
The equation is now in the form \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = -6 \), and \( c = 0 \).
- *Finding Roots*:
Set \( y = 0 \) and solve:
\[ 3x^2 - 6x = 0 \]
\[ 3x(x - 2) = 0 \]
Roots are \( x = 0 \) and \( x = 2 \).
- *Calculating Vertex*:
Using the vertex formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{-6}{2 \times 3} = 1 \]
Substitute back to find \( y \):
\[ y = 3(1)^2 - 6(1) = 3 - 6 = -3 \]
The vertex is at \( (1, -3) \).
**Conclusion**:
The quadratic equation has been converted to standard form, and its roots and vertex have been successfully calculated.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F532d2df1-6bef-42be-bbaa-e1703c54e78f%2Fb846e0a7-01f6-485a-831c-c07ff8e835b7%2Fro23zf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Quadratic Conversion and Analysis**
**Objective**: Convert the quadratic into standard form by detailing the steps involved in changing forms. Then, identify the roots and vertex of the given quadratic equation.
**Problem Statement**:
Given the quadratic equation:
\[ y = 3x(x - 2) \]
**Instructions**:
1. **Expand the Equation**: Distribute the terms to convert the equation into standard form.
2. **Standard Form**: Express the quadratic in the standard form of \( ax^2 + bx + c \).
3. **Identify Roots**: Solve for the values of \( x \) where \( y = 0 \).
4. **Find the Vertex**: Calculate the vertex of the parabola using the standard form.
**Steps to Solution**:
- *Expanding the Quadratic*:
\[ y = 3x^2 - 6x \]
- *Standard Form*:
The equation is now in the form \( ax^2 + bx + c = 0 \), where \( a = 3 \), \( b = -6 \), and \( c = 0 \).
- *Finding Roots*:
Set \( y = 0 \) and solve:
\[ 3x^2 - 6x = 0 \]
\[ 3x(x - 2) = 0 \]
Roots are \( x = 0 \) and \( x = 2 \).
- *Calculating Vertex*:
Using the vertex formula \( x = -\frac{b}{2a} \):
\[ x = -\frac{-6}{2 \times 3} = 1 \]
Substitute back to find \( y \):
\[ y = 3(1)^2 - 6(1) = 3 - 6 = -3 \]
The vertex is at \( (1, -3) \).
**Conclusion**:
The quadratic equation has been converted to standard form, and its roots and vertex have been successfully calculated.
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