Question

1. consider the two functions below. circle the function below that is linear. write the equation for the linear function.

function a
graph of a line on a coordinate plane with x-axis from -4 to 4 and y-axis from -2 to 4
function b

x	1	2	3	4	5
y	1	8	27	64	125

equation is ____________

Explanation:

Step1: Identify linear function

A linear function has a constant rate of change (slope) and graphs as a straight line. Function A is a straight line (from the graph), Function B: check differences. For Function B, \( y \) values are \( 1 = 1^3 \), \( 8 = 2^3 \), \( 27 = 3^3 \), \( 64 = 4^3 \), \( 125 = 5^3 \), so it's \( y = x^3 \) (non - linear). So Function A is linear.

Step2: Find slope of Function A

From the graph, pick two points. Let's take \( (0,1) \) (wait, no, looking at the graph: when \( x = 0 \), \( y = 1 \)? Wait, no, let's re - examine. Wait, the line passes through \( (0,1) \)? Wait, no, let's take two points. Let's take \( (- 1,0) \) and \( (0,1) \)? Wait, no, the line in Function A: when \( x=-2 \), \( y = - 1 \)? Wait, maybe better to use two clear points. Let's take \( (0,1) \) and \( (2,3) \)? Wait, no, looking at the grid: the line passes through \( (0,1) \) and \( (2,3) \)? Wait, no, let's calculate slope. Let's take two points: when \( x = 0 \), \( y = 1 \)? Wait, no, maybe \( (0,1) \) and \( (1,2) \)? Wait, the slope \( m=\frac{y_2 - y_1}{x_2 - x_1}\). Let's take \( (0,1) \) and \( (1,2) \): \( m=\frac{2 - 1}{1 - 0}=1 \). Wait, another way: the line passes through \( (0,1) \) and \( (-1,0) \): \( m=\frac{0 - 1}{-1 - 0}=1 \). Wait, or \( (2,3) \) and \( (1,2) \): \( m = 1 \). And the y - intercept \( b \) is 1? Wait, no, wait the graph: when \( x = 0 \), \( y = 1 \)? Wait, no, maybe I made a mistake. Wait, looking at the graph, the line passes through \( (0,1) \) and has a slope of 1? Wait, no, let's check again. Wait, the line in Function A: when \( x=-2 \), \( y=-1 \); \( x = 0 \), \( y = 1 \); \( x = 2 \), \( y = 3 \). So the slope \( m=\frac{1-(-1)}{0 - (-2)}=\frac{2}{2}=1 \). And the y - intercept \( b = 1 \)? Wait, no, when \( x = 0 \), \( y = 1 \), so the equation is \( y=mx + b \), \( m = 1 \), \( b = 1 \)? Wait, no, wait \( x = 0 \), \( y = 1 \); \( x = 1 \), \( y = 2 \); \( x = 2 \), \( y = 3 \). So the slope is \( \frac{2 - 1}{1 - 0}=1 \), and y - intercept \( b = 1 \). Wait, but let's check with \( x=-1 \), \( y = 0 \): \( y=1\times(-1)+1 = 0 \), which matches. So the equation is \( y=x + 1 \)? Wait, no, wait the graph: maybe I misread. Wait, the line in Function A: when \( x = 0 \), \( y = 1 \), and slope 1. So the equation is \( y=x + 1 \)? Wait, but let's check another point. When \( x = 2 \), \( y=3 \), and \( 2 + 1=3 \), which matches. When \( x=-2 \), \( y=-2 + 1=-1 \), which matches the graph (since at \( x=-2 \), \( y=-1 \)). So the equation is \( y=x + 1 \)? Wait, no, wait the original graph: maybe the y - intercept is 1? Wait, but let's re - check. Alternatively, maybe the line passes through \( (0,1) \) and has slope 1. So the equation is \( y=x + 1 \). Wait, but let's confirm with the two - point formula. Let's take two points \( (x_1,y_1)=(0,1) \) and \( (x_2,y_2)=(2,3) \). Then \( m=\frac{3 - 1}{2 - 0}=\frac{2}{2}=1 \). Then using point - slope form \( y - y_1=m(x - x_1) \), \( y - 1=1\times(x - 0) \), so \( y=x + 1 \).

Answer:

The linear function is Function A. The equation of the linear function is \( y = x+1 \) (Wait, no, wait I think I made a mistake. Wait, looking at the graph again: when \( x = 0 \), \( y = 1 \)? Wait, no, maybe the y - intercept is 1? Wait, no, let's check the line again. Wait, the line in Function A: when \( x=-1 \), \( y = 0 \); \( x = 0 \), \( y = 1 \); \( x = 1 \), \( y = 2 \). So slope is 1, y - intercept 1. So equation \( y=x + 1 \).