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Chapter 4: Problem 13

Are the two functions the same function? $$ r(x)=5(x-2)+3 \text { and } s(x)=5 x+7 $$

Short Answer

Expert verified
Explain your reasoning. Answer: No, the functions r(x) and s(x) are not the same. After simplifying and expanding r(x), we find that r(x) = 5x - 7. When comparing this to s(x) = 5x + 7, we can see that the only difference between these two functions is the sign of the constant term (r(x) has -7, while s(x) has +7). Therefore, they are not the same function.

Step by step solution

01

Expanding the function r(x)

First, we need to expand the function r(x) by distributing the term 5 to (x-2) in the parenthesis. This can be done using the distributive property: $$ r(x) = 5(x-2)+3 $$ $$ r(x) = 5x - 10 + 3 $$
02

Simplifying the function r(x)

Next, simplify the function r(x) by combining the constants -10 and 3: $$ r(x) = 5x - 7 $$
03

Comparing r(x) and s(x)

Now, we'll compare our simplified r(x) with the given function s(x) to see if they are the same function: $$ r(x) = 5x - 7 $$ $$ s(x) = 5x + 7 $$ At this point, we notice that the only difference between these two functions is the sign of the constant term: r(x) has -7, while s(x) has +7. Since the functions differ in the constant term, they are not the same function. In conclusion, the two functions r(x) and s(x) are not the same function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distributive Property
The distributive property is a fundamental algebraic principle that allows us to multiply a single term across terms inside a set of parentheses. It's a handy tool when dealing with expressions involving parentheses, as it helps simplify them. For example, consider the expression \(5(x-2)\). To simplify it, we multiply 5 by each term inside the parenthesis:
  • Multiply 5 by \(x\) to get \(5x\).
  • Multiply 5 by -2 to get -10.
This results in the expression \(5x - 10\).
Using the distributive property correctly helps us in rewriting expressions and is particularly useful when simplifying or factoring expressions in algebra. Understanding this principle is essential for mastering more complex algebraic functions.
Function Comparison
Function comparison is an important skill in algebra, enabling students to determine if two functions are equivalent. To compare functions, we focus on their form, expression, or output. In the example provided with functions \(r(x)=5(x-2)+3\) and \(s(x)=5x+7\), our task is to determine if they represent the same curve in a graph.
  • Firstly, simplify each function as much as possible.
  • In the solved exercise, \(r(x)\) was expanded to \(5x - 10 + 3\) and simplified to \(5x - 7\).
  • Next, compare \(5x - 7\) with \(5x + 7\).
Notice how the expressions differ only in the constant term: -7 and +7, respectively.
Different constant terms indicate that the functions are not equivalent, as they affect the output values of the function. Comparing functions helps in understanding their behavior and properties in various contexts.
Constant Terms
Constant terms in algebraic functions are the numbers that do not change with the variable; they remain consistent throughout the function's domain. They can greatly influence the position or height of a function on a graph. In the functions \(r(x) = 5x - 7\) and \(s(x) = 5x + 7\), the constant terms are -7 and +7.
  • Constant terms determine where a line crosses the y-axis when graphed, which can be seen as the function's offset.
  • The sign and value of a constant term can change the function's range of outputs significantly.
Understanding how constant terms work can help students better graph functions and predict changes based on alterations to these constants. In comparing functions, a difference in constant terms鈥攍ike the negative and positive 7 in the provided example鈥攎eans the two functions will have different outputs for the same inputs, confirming they are not the same function.

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Most popular questions from this chapter

Table 4.11 shows values of \(v\) and the expressions \(12-3 v\) and \(-3+2 v\). For which values of \(v\) in the table is (a) \(12-3 v<-3+2 v ?\) (b) \(12-3 v>-3+2 v ?\) (c) \(12-3 v=-3+2 v ?\) Table 4.11 $$ \begin{array}{c|c|c|c|c|c|c|c} \hline v & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 12-3 v & 12 & 9 & 6 & 3 & 0 & -3 & -6 \\ \hline-3+2 v & -3 & -1 & 1 & 3 & 5 & 7 & 9 \\ \hline \end{array} $$

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