Let's calculate the values of \( f(x) = 4x + 3 \) and \( g(x) = 4x^2 \) for the given values of \( x \):
Table of Values:
| x | f(x) = 4x + 3 | g(x) = 4x^2 | |-----|------------------|----------------| | 0 | \( 4(0) + 3 = 3 \) | \( 4(0)^2 = 0 \) | | 1 | \( 4(1) + 3 = 7 \) | \( 4(1)^2 = 4 \) | | 2 | \( 4(2) + 3 = 11 \) | \( 4(2)^2 = 16 \) | | 3 | \( 4(3) + 3 = 15 \) | \( 4(3)^2 = 36 \) |
Completed Table:
| x | f(x) | g(x) | |-----|-------|-------| | 0 | 3 | 0 | | 1 | 7 | 4 | | 2 | 11 | 16 | | 3 | 15 | 36 |
Analysis of Growth:
Both functions \( f(x) = 4x + 3 \) (a linear function) and \( g(x) = 4x^2 \) (a quadratic function) do grow as \( x \) increases. However, a linear function grows at a constant rate, while a quadratic function grows at an increasing rate.
As \( x \) gets larger, \( g(x) = 4x^2 \) will eventually exceed \( f(x) = 4x + 3 \). Specifically, quadratic functions will always surpass linear functions as \( x \) approaches infinity.
In this case:
- At \( x = 0 \), \( f(0) = 3 \) and \( g(0) = 0 \)
- At \( x = 1 \), \( f(1) = 7 \) and \( g(1) = 4 \)
- At \( x = 2 \), \( f(2) = 11 \) and \( g(2) = 16 \) (here \( g(x) \) exceeds \( f(x) \))
- By \( x = 3 \), \( g(3) = 36 \), which is much greater than \( f(3) = 15 \)
Thus, the function \( g(x) \) eventually exceeds \( f(x) \).