Question

unit two -- miniquiz
linear vs. exponential functions
name:
date:

1. examine the tables below: first identify if the functio

and then write the equation to model the data. (each
exponential or linear? (circle one)
equation:

Explanation:

Step1: Check if it is linear

For a linear function, the difference in \(y\)-values (\(\Delta y\)) should be constant when \(x\) changes by a constant \(\Delta x\). Here, \(\Delta x = 1\) (from \(x=-2\) to \(x = - 1\), \(x=-1\) to \(x = 0\), etc.).
\(\Delta y\) from \(x=-2\) to \(x=-1\): \(18 - 54=-36\)
\(\Delta y\) from \(x=-1\) to \(x = 0\): \(6 - 18=-12\)
Since \(-36
eq-12\), it is not linear.

Step2: Check if it is exponential

For an exponential function \(y = a\cdot b^{x}\), when \(x = 0\), \(y=a\). From the table, when \(x = 0\), \(f(x)=6\), so \(a = 6\).
When \(x = 1\), \(y=2\). Substitute \(a = 6\) and \(x = 1\), \(y = 2\) into \(y=a\cdot b^{x}\), we get \(2=6\cdot b^{1}\), then \(b=\frac{2}{6}=\frac{1}{3}\)
Check for \(x=-1\): \(y = 6\cdot(\frac{1}{3})^{-1}=6\times3 = 18\)
Check for \(x=-2\): \(y = 6\cdot(\frac{1}{3})^{-2}=6\times9 = 54\)

Answer:

Exponential
Equation: \(f(x)=6\cdot(\frac{1}{3})^{x}\)