2) $f(x) = 5 + X$3) Complete The Table Using The Function $f(x) = 5 + X$:$\[ \begin{tabular}{|c|c|c|c|c|c|} \hline $x$ & -3 & -1 & 3 & 5 & 7 \\ \hline $f(x)$ & 2 & 4 & 8 & 10 & 12

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Introduction

In mathematics, a linear function is a function that can be written in the form of f(x)=mx+bf(x) = mx + b, where mm is the slope and bb is the y-intercept. In this article, we will be working with a linear function of the form f(x)=5+xf(x) = 5 + x. We will use this function to complete a table with given values of xx.

Understanding the Function

The given function is f(x)=5+xf(x) = 5 + x. This means that for every value of xx, we add 5 to it to get the corresponding value of f(x)f(x). For example, if x=2x = 2, then f(x)=5+2=7f(x) = 5 + 2 = 7.

Completing the Table

We are given a table with values of xx and corresponding values of f(x)f(x). However, the values of f(x)f(x) are not calculated using the function f(x)=5+xf(x) = 5 + x. We will use this function to calculate the correct values of f(x)f(x) for each value of xx.

x f(x)
-3 2
-1 4
3 8
5 10
7 12

To complete the table, we will substitute each value of xx into the function f(x)=5+xf(x) = 5 + x and calculate the corresponding value of f(x)f(x).

Calculating f(x) for x = -3

f(x)=5+xf(x) = 5 + x f(βˆ’3)=5+(βˆ’3)f(-3) = 5 + (-3) f(βˆ’3)=5βˆ’3f(-3) = 5 - 3 f(βˆ’3)=2f(-3) = 2

Calculating f(x) for x = -1

f(x)=5+xf(x) = 5 + x f(βˆ’1)=5+(βˆ’1)f(-1) = 5 + (-1) f(βˆ’1)=5βˆ’1f(-1) = 5 - 1 f(βˆ’1)=4f(-1) = 4

Calculating f(x) for x = 3

f(x)=5+xf(x) = 5 + x f(3)=5+3f(3) = 5 + 3 f(3)=5+3f(3) = 5 + 3 f(3)=8f(3) = 8

Calculating f(x) for x = 5

f(x)=5+xf(x) = 5 + x f(5)=5+5f(5) = 5 + 5 f(5)=5+5f(5) = 5 + 5 f(5)=10f(5) = 10

Calculating f(x) for x = 7

f(x)=5+xf(x) = 5 + x f(7)=5+7f(7) = 5 + 7 f(7)=5+7f(7) = 5 + 7 f(7)=12f(7) = 12

Completed Table

x f(x)
-3 2
-1 4
3 8
5 10
7 12

Discussion

In this article, we used the linear function f(x)=5+xf(x) = 5 + x to complete a table with given values of xx. We calculated the corresponding values of f(x)f(x) for each value of xx using the function. This demonstrates how to use a linear function to complete a table with given values.

Conclusion

In conclusion, completing a table using a linear function involves substituting each value of xx into the function and calculating the corresponding value of f(x)f(x). This is a useful skill in mathematics, as it allows us to work with linear functions and complete tables with given values.

Example Problems

  1. Complete the table using the function f(x)=2x+1f(x) = 2x + 1:
x f(x)
-2 3
0 1
4 9
6 13
  1. Complete the table using the function f(x)=βˆ’3x+2f(x) = -3x + 2:
x f(x)
-1 5
2 -4
5 -13
8 -22

Answer Key

1. x f(x)
-2 3
0 1
4 9
6 13

f(βˆ’2)=2(βˆ’2)+1=βˆ’3+1=βˆ’2+3=1f(-2) = 2(-2) + 1 = -3 + 1 = -2 + 3 = 1 f(0)=2(0)+1=0+1=1f(0) = 2(0) + 1 = 0 + 1 = 1 f(4)=2(4)+1=8+1=9f(4) = 2(4) + 1 = 8 + 1 = 9 f(6)=2(6)+1=12+1=13f(6) = 2(6) + 1 = 12 + 1 = 13

2. x f(x)
-1 5
2 -4
5 -13
8 -22

Introduction

In our previous article, we discussed how to complete a table using a linear function. We used the function f(x)=5+xf(x) = 5 + x to complete a table with given values of xx. In this article, we will answer some frequently asked questions about completing tables using linear functions.

Q: What is a linear function?

A linear function is a function that can be written in the form of f(x)=mx+bf(x) = mx + b, where mm is the slope and bb is the y-intercept.

Q: How do I know if a function is linear?

To determine if a function is linear, you can check if it can be written in the form of f(x)=mx+bf(x) = mx + b. If it can, then it is a linear function.

Q: What is the slope of a linear function?

The slope of a linear function is the coefficient of xx in the function. For example, in the function f(x)=2x+1f(x) = 2x + 1, the slope is 2.

Q: What is the y-intercept of a linear function?

The y-intercept of a linear function is the value of bb in the function. For example, in the function f(x)=2x+1f(x) = 2x + 1, the y-intercept is 1.

Q: How do I complete a table using a linear function?

To complete a table using a linear function, you need to substitute each value of xx into the function and calculate the corresponding value of f(x)f(x).

Q: What if I have a table with values of xx and corresponding values of f(x)f(x), but the values of f(x)f(x) are not calculated using the function?

In this case, you can use the function to calculate the correct values of f(x)f(x) for each value of xx.

Q: Can I use a linear function to complete a table with negative values of xx?

Yes, you can use a linear function to complete a table with negative values of xx. The function will still work as long as the values of xx are in the domain of the function.

Q: Can I use a linear function to complete a table with decimal values of xx?

Yes, you can use a linear function to complete a table with decimal values of xx. The function will still work as long as the values of xx are in the domain of the function.

Q: How do I know if a table is completed correctly using a linear function?

To check if a table is completed correctly using a linear function, you can plug in the values of xx and f(x)f(x) into the function and see if the equation holds true.

Q: What if I make a mistake when completing a table using a linear function?

If you make a mistake when completing a table using a linear function, you can go back and recheck your work. You can also use a calculator or a computer program to help you complete the table.

Conclusion

In conclusion, completing a table using a linear function is a useful skill in mathematics. By understanding the concept of linear functions and how to complete tables using them, you can solve a wide range of problems in mathematics and other fields.

Example Problems

  1. Complete the table using the function f(x)=2x+1f(x) = 2x + 1:
x f(x)
-2 3
0 1
4 9
6 13
  1. Complete the table using the function f(x)=βˆ’3x+2f(x) = -3x + 2:
x f(x)
-1 5
2 -4
5 -13
8 -22

Answer Key

1. x f(x)
-2 3
0 1
4 9
6 13

f(βˆ’2)=2(βˆ’2)+1=βˆ’4+1=βˆ’3+3=0f(-2) = 2(-2) + 1 = -4 + 1 = -3 + 3 = 0 f(0)=2(0)+1=0+1=1f(0) = 2(0) + 1 = 0 + 1 = 1 f(4)=2(4)+1=8+1=9f(4) = 2(4) + 1 = 8 + 1 = 9 f(6)=2(6)+1=12+1=13f(6) = 2(6) + 1 = 12 + 1 = 13

2. x f(x)
-1 5
2 -4
5 -13
8 -22

f(βˆ’1)=βˆ’3(βˆ’1)+2=3+2=5f(-1) = -3(-1) + 2 = 3 + 2 = 5 f(2)=βˆ’3(2)+2=βˆ’6+2=βˆ’4f(2) = -3(2) + 2 = -6 + 2 = -4 f(5)=βˆ’3(5)+2=βˆ’15+2=βˆ’13f(5) = -3(5) + 2 = -15 + 2 = -13 f(8)=βˆ’3(8)+2=βˆ’24+2=βˆ’22f(8) = -3(8) + 2 = -24 + 2 = -22