Question

1. a function has selected values given in the table below:

$x$	0	1	2	3	4
$f(x)$	81	108	144	192	256

(a) explain how you can tell from the table that this function cannot be linear. 2 points
(b) an exponential function of the form $f(x)=a(b)^x$ will correctly model the data in the table. find values for $a$ and $b$. justify how you found both values. 4 points

Explanation:

Part (a)

Step1: Recall linear function property

A linear function has a constant rate of change (slope), meaning the difference between consecutive \( f(x) \) values (when \( x \) increases by 1) should be constant.

Step2: Calculate differences

For \( x = 0 \) to \( x = 1 \): \( 108 - 81 = 27 \)
For \( x = 1 \) to \( x = 2 \): \( 144 - 108 = 36 \)
For \( x = 2 \) to \( x = 3 \): \( 192 - 144 = 48 \)
For \( x = 3 \) to \( x = 4 \): \( 256 - 192 = 64 \)
The differences (27, 36, 48, 64) are not constant, so the function cannot be linear.

Step1: Find \( a \) using \( x = 0 \)

The exponential function is \( f(x)=a(b)^x \). When \( x = 0 \), \( f(0)=a(b)^0=a(1)=a \). From the table, \( f(0) = 81 \), so \( a = 81 \).

Step2: Find \( b \) using another point

Use \( x = 1 \), \( f(1)=108 \), and \( a = 81 \). Substitute into \( f(x)=a(b)^x \): \( 108 = 81(b)^1 \). Solve for \( b \): \( b=\frac{108}{81}=\frac{4}{3} \). We can verify with other points (e.g., \( x = 2 \): \( f(2)=81(\frac{4}{3})^2 = 81\times\frac{16}{9}=144 \), which matches the table).

Answer:

To determine a function is linear, the difference between consecutive \( f(x) \) values (for \( x \) increasing by 1) must be constant. Here, the differences between \( f(x) \) values are \( 108 - 81 = 27 \), \( 144 - 108 = 36 \), \( 192 - 144 = 48 \), \( 256 - 192 = 64 \). Since these differences (27, 36, 48, 64) are not constant, the function cannot be linear.

Part (b)