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(a).
To graph: Quadratic function with its vertex, axis of symmetry,
(a).
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Answer to Problem 23RE
Function using the vertex and the
Explanation of Solution
Given:
Calculation:
The quadratic function
So,
The graph of a quadratic function
Here, function has a
So the graph opens up.
To find the vertex,
The coordinates of the vertex of a quadratic function of the form
Evaluating
Evaluating
Thus, the vertex is at
The axis of symmetry is the line where
Thus,
To find
Here,
Therefore, the graph has the
Find the discriminant
In case discriminant
Therefore, the graph of the function crosses the
To find out the
The
Substitute,
Function using the vertex and the
Conclusion:
Quadratic function with its vertex is
(b).
Domain and range of the function.
(b).
![Check Mark](/static/check-mark.png)
Answer to Problem 23RE
Domain is the interval
Explanation of Solution
Given:
Calculation:
Find the domain and range of the function, the domain of
From the graph, find the domain is the interval
Conclusion:
Thus, domain is the interval
(c).
The function is increasing and where it is decreasing.
(c).
![Check Mark](/static/check-mark.png)
Answer to Problem 23RE
The function
Explanation of Solution
Given:
Calculation:
Interval where the function
Also, find where the function is increasing.
From the graph, the function
Conclusion:
Thus, the function
Chapter 3 Solutions
Precalculus
Additional Math Textbook Solutions
Precalculus Enhanced with Graphing Utilities (7th Edition)
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (4th Edition)
Precalculus
Thomas' Calculus: Early Transcendentals (14th Edition)
Single Variable Calculus: Early Transcendentals (2nd Edition) - Standalone book
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- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning
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