Alright class, let's examine the relationship between the input [tex]\( x \)[/tex] and the output [tex]\( y \)[/tex] step-by-step with the data provided.
We have the following pairs of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
-1 & -4 \\
0 & -1 \\
1 & 2 \\
2 & 5 \\
3 & 8 \\
4 & 11 \\
5 & 14 \\
\hline
\end{array}
\][/tex]
To identify the relationship, let's look for a pattern.
1. Determine the changes in [tex]\( y \)[/tex] (Δy) for each unit increase in [tex]\( x \)[/tex]:
[tex]\[
\begin{aligned}
2 - (-1) &= 3, \\
5 - 2 &= 3, \\
8 - 5 &= 3, \\
11 - 8 &= 3, \\
14 - 11 &= 3. \\
\end{aligned}
\][/tex]
We see that each time [tex]\( x \)[/tex] increases by 1, [tex]\( y \)[/tex] increases by 3. This constant rate of change indicates a linear relationship.
2. Formulate the linear equation in the form [tex]\( y = mx + b \)[/tex]:
Given that the slope [tex]\( m \)[/tex] (change in [tex]\( y \)[/tex] per unit change in [tex]\( x \)[/tex]) is 3 and one of the points [tex]\( (0, -1) \)[/tex], we can use these to find the y-intercept [tex]\( b \)[/tex].
We know from the form [tex]\( y = mx + b \)[/tex]:
[tex]\[
\begin{aligned}
y &= mx + b, \\
-1 &= 3(0) + b, \\
b &= -1.
\end{aligned}
\][/tex]
So, our equation becomes [tex]\( y = 3x - 1 \)[/tex].
3. Verification:
Let's verify this equation using a few points from our table:
[tex]\[\begin{aligned}
x &= -1: & y &= 3(-1) - 1 = -3 - 1 = -4,\\
x &= 0: & y &= 3(0) - 1 = 0 - 1 = -1,\\
x &= 1: & y &= 3(1) - 1 = 3 - 1 = 2,\\
x &= 2: & y &= 3(2) - 1 = 6 - 1 = 5.\\
\end{aligned}\][/tex]
Each point satisfies the equation [tex]\( y = 3x - 1 \)[/tex].
Thus, the relationship between the input [tex]\( x \)[/tex] and the output [tex]\( y \)[/tex] is given by the linear equation:
[tex]\[
y = 3x - 1
\][/tex]
This is the equation that describes the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
