Mrs. Jackson gives the table below to her students.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
4 & 26 \\
\hline
5 & 23 \\
\hline
6 & [tex]$a$[/tex] \\
\hline
\end{tabular}

In order for the function to be linear, what must [tex]$a$[/tex] be and why?

A. [tex]$a=22$[/tex] because the rate of change is 1.
B. [tex]$a=20$[/tex] because the rate of change is 3.
C. [tex]$a=22$[/tex] because the rate of change is -1.
D. [tex]$a=20$[/tex] because the rate of change is -3.



Answer :

To determine the value of [tex]\(a\)[/tex] in the table for a linear function, we need to ensure that the rate of change, or the slope, between any two pairs of points is consistent.

The provided points in the table are:
[tex]\[ (4, 26) \quad \text{and} \quad (5, 23) \][/tex]

First, we calculate the rate of change (slope) between these two points:
[tex]\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{23 - 26}{5 - 4} = \frac{-3}{1} = -3 \][/tex]

This slope must be consistent between all pairs of points for the function to be linear. Therefore, the rate of change (slope) between [tex]\( (5, 23) \)[/tex] and [tex]\( (6, a) \)[/tex] should also be [tex]\(-3\)[/tex].

Using the slope formula with these points:
[tex]\[ -3 = \frac{a - 23}{6 - 5} = \frac{a - 23}{1} \][/tex]

Solving for [tex]\(a\)[/tex]:
[tex]\[ -3 = a - 23 \][/tex]
[tex]\[ a = 23 - 3 \][/tex]
[tex]\[ a = 20 \][/tex]

Thus, the value of [tex]\(a\)[/tex] must be [tex]\(20\)[/tex] to ensure the function remains linear.

Therefore, the correct answer is:
[tex]\[ a = 20 \text{ because the rate of change is } -3. \][/tex]