Question

mrs. jackson gives the table to her students.

$x$	$y$
5	23
6	$a$

in order for the function to be linear, what must $a$ be and why?

• $a = 22$ because the rate of change is $-1$.
• $a = 22$ because the rate of change is $1$.
• $a = 20$ because the rate of change is $-3$.
• $a = 20$ because the rate of change is $3$.

Explanation:

Step1: Calculate rate of change between x=4 and x=5

The rate of change (slope) between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\frac{y_2 - y_1}{x_2 - x_1}\). For \(x_1 = 4,y_1 = 26\) and \(x_2 = 5,y_2 = 23\), the rate of change is \(\frac{23 - 26}{5 - 4}=\frac{-3}{1}=-3\)? Wait, no, wait. Wait, let's recalculate. Wait, \(23 - 26=-3\), \(5 - 4 = 1\), so slope is \(-3\)? Wait, no, wait, let's check again. Wait, when \(x\) increases by 1 (from 4 to 5), \(y\) changes from 26 to 23, so the change in \(y\) is \(23 - 26=-3\). So the rate of change (slope) is \(-3\) per unit increase in \(x\). Now, for the function to be linear, the rate of change between \(x = 5\) and \(x = 6\) should be the same. So when \(x\) goes from 5 to 6 (increase by 1), \(y\) should decrease by 3 (since rate of change is -3). So \(y\) at \(x = 6\) (which is \(a\)) should be \(23+(-3)=20\). Let's verify the rate of change between \(x = 4\) and \(x = 5\): \(\frac{23 - 26}{5 - 4}=-3\). Between \(x = 5\) and \(x = 6\): \(\frac{a - 23}{6 - 5}=\frac{a - 23}{1}\). For linear function, this should equal -3. So \(a - 23=-3\), so \(a=23 - 3 = 20\). So the rate of change is -3, and \(a = 20\).

Step2: Confirm the rate of change and find a

Since the rate of change (slope) of a linear function is constant. The change in \(x\) from 4 to 5 is \(5 - 4 = 1\), change in \(y\) is \(23 - 26=-3\). So the slope \(m=-3\). For \(x\) from 5 to 6 (change in \(x\) is \(6 - 5 = 1\)), the change in \(y\) should also be \(-3\). So \(a=23+(-3)=20\). So the correct option is \(a = 20\) because the rate of change is -3.

Answer:

\(a = 20\) because the rate of change is \(-3\) (the option: \(a = 20\) because the rate of change is \(-3\)).