Question

answer the questions below to determine what kind of function is depicted in the table below.

$x$	1	2	3	4	5
$f(x)$	$-3$	$43$	$181$	$411$	$733$

this function is because .

Explanation:

Step1: Calculate first differences

Find the differences between consecutive \( f(x) \) values:
\( 43 - (-3) = 46 \), \( 181 - 43 = 138 \), \( 411 - 181 = 230 \), \( 733 - 411 = 322 \).

Step2: Calculate second differences

Find the differences of the first differences:
\( 138 - 46 = 92 \), \( 230 - 138 = 92 \), \( 322 - 230 = 92 \).

Step3: Analyze the differences

The second differences are constant (\( 92 \)). For a function, if the \( n \)-th differences are constant, it is a polynomial of degree \( n \). Here, second differences are constant, so it is a quadratic (degree 2) function? Wait, no—wait, second differences constant implies degree 2? Wait, no, let's check again. Wait, first differences: 46, 138, 230, 322. Second differences: 92, 92, 92. So second differences are constant. So the function is a quadratic (degree 2) function? Wait, no, actually, for a polynomial function, the \( k \)-th differences being constant means the function is of degree \( k \). So if second differences are constant, it's a quadratic (degree 2) function? Wait, but let's test with a cubic? Wait, no, let's check the pattern. Wait, maybe I made a mistake. Wait, let's assume the function is a cubic: \( f(x) = ax^3 + bx^2 + cx + d \). Let's plug in \( x = 1 \): \( a + b + c + d = -3 \). \( x = 2 \): \( 8a + 4b + 2c + d = 43 \). \( x = 3 \): \( 27a + 9b + 3c + d = 181 \). \( x = 4 \): \( 64a + 16b + 4c + d = 411 \). Subtract first from second: \( 7a + 3b + c = 46 \) (eq1). Subtract second from third: \( 19a + 5b + c = 138 \) (eq2). Subtract third from fourth: \( 37a + 7b + c = 230 \) (eq3). Now subtract eq1 from eq2: \( 12a + 2b = 92 \) → \( 6a + b = 46 \) (eq4). Subtract eq2 from eq3: \( 18a + 2b = 92 \) → \( 9a + b = 46 \) (eq5). Subtract eq4 from eq5: \( 3a = 0 \) → \( a = 0 \). Wait, that can't be. Wait, no, that would make it quadratic. But if \( a = 0 \), then it's quadratic. But when \( a = 0 \), from eq4: \( 0 + b = 46 \) → \( b = 46 \). Then from eq1: \( 0 + 3*46 + c = 46 \) → \( 138 + c = 46 \) → \( c = -92 \). Then from first equation: \( 0 + 46 - 92 + d = -3 \) → \( -46 + d = -3 \) → \( d = 43 \). So \( f(x) = 46x^2 - 92x + 43 \). Let's test \( x = 3 \): \( 46*9 - 92*3 + 43 = 414 - 276 + 43 = 181 \). Correct. \( x = 4 \): \( 46*16 - 92*4 + 43 = 736 - 368 + 43 = 411 \). Correct. \( x = 5 \): \( 46*25 - 92*5 + 43 = 1150 - 460 + 43 = 733 \). Correct. So the function is quadratic (degree 2), because second differences are constant. So the function is a quadratic (polynomial of degree 2) function, so it is a quadratic (or second-degree polynomial) function, and the reason is that the second differences of the function values are constant.

Wait, but earlier I thought maybe cubic, but the calculation shows it's quadratic. So the key is: for a function represented by a table, we check the differences (first, second, third, etc.). If the \( n \)-th differences are constant, the function is a polynomial of degree \( n \). Here, second differences are constant (92), so it's a quadratic (degree 2) function.

So the function is a quadratic (or second - degree polynomial) function because the second differences of the \( f(x) \) values (when \( x \) increases by 1 each time) are constant.

Answer:

This function is a \(\boldsymbol{\text{quadratic (second - degree polynomial)}}\) function because the second differences of the \( f(x) \) values (for consecutive integer \( x \)-values) are constant (all equal to 92).