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,...
Related questions
Question
![**Problem:**
If \( f(x) = ax^2 + bx + c \) has zeros \( x = 15 \) and \( x = 6 \), what zeros does \( f(x+1) \) have?
**Solution:**
Given the quadratic function \( f(x) = ax^2 + bx + c \) with zeros at \( x = 15 \) and \( x = 6 \), we want to find the zeros of the function \( f(x+1) \).
To find this, we replace \( x \) with \( x-1 \) to account for the shift in the variable:
1. Original zeros: \( x = 15 \) and \( x = 6 \).
2. Solving for \( x \) in \( f(x+1) = 0 \), substitute:
\[
f(x+1) \Rightarrow f((x+1)-1) = 0
\]
3. New zeros are \( x+1 = 15 \) and \( x+1 = 6 \).
4. Solve for \( x \):
- \( x+1 = 15 \) implies \( x = 14 \).
- \( x+1 = 6 \) implies \( x = 5 \).
Thus, the zeros of \( f(x+1) \) are \( x = 14 \) and \( x = 5 \).
**Answer Box:**
\[ x = \]
In the blank space, the user would fill in with the calculated zero values: \( 14 \) or \( 5 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf99ef9a-11bc-4c83-b49c-a6b52894d845%2Fd160b406-5f35-415a-a477-2934c29c519c%2Fkaxys0e_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
If \( f(x) = ax^2 + bx + c \) has zeros \( x = 15 \) and \( x = 6 \), what zeros does \( f(x+1) \) have?
**Solution:**
Given the quadratic function \( f(x) = ax^2 + bx + c \) with zeros at \( x = 15 \) and \( x = 6 \), we want to find the zeros of the function \( f(x+1) \).
To find this, we replace \( x \) with \( x-1 \) to account for the shift in the variable:
1. Original zeros: \( x = 15 \) and \( x = 6 \).
2. Solving for \( x \) in \( f(x+1) = 0 \), substitute:
\[
f(x+1) \Rightarrow f((x+1)-1) = 0
\]
3. New zeros are \( x+1 = 15 \) and \( x+1 = 6 \).
4. Solve for \( x \):
- \( x+1 = 15 \) implies \( x = 14 \).
- \( x+1 = 6 \) implies \( x = 5 \).
Thus, the zeros of \( f(x+1) \) are \( x = 14 \) and \( x = 5 \).
**Answer Box:**
\[ x = \]
In the blank space, the user would fill in with the calculated zero values: \( 14 \) or \( 5 \).
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