Question

unit two - miniquiz
linear vs. exponential functions
retake
name:
score: /14
date:
period:

1. examine the tables below: first identify if the function represents a linear or exponential function and then write the equation to model the data. (each @ 3 points)

a.
exponential or linear? (circle one)
equation:
b.
exponential or linear? (circle one)
equation:

Explanation:

Step1: Check for linearity in part a

For a linear function, the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\) is constant.
Take \((x_1,y_1)=(-2,-7)\) and \((x_2,y_2)=(-1,-4)\): \(m=\frac{-4-(-7)}{-1 - (-2)}=\frac{-4 + 7}{-1+2}=3\)
Take \((x_1,y_1)=(-1,-4)\) and \((x_2,y_2)=(0,-1)\): \(m=\frac{-1-(-4)}{0-(-1)}=\frac{-1 + 4}{0 + 1}=3\)
Take \((x_1,y_1)=(0,-1)\) and \((x_2,y_2)=(1,2)\): \(m=\frac{2-(-1)}{1-0}=\frac{2 + 1}{1}=3\)
Since the slope is constant, it is a linear function.
The equation of a line is \(y=mx + b\). Using the point \((0,-1)\), \(b=-1\) (y - intercept). So the equation is \(y = 3x-1\)

Step2: Check for linearity or exponential in part b

For a linear function, the slope \(m\) should be constant.
\(m_1=\frac{14 - 7}{1-0}=7\), \(m_2=\frac{28 - 14}{2 - 1}=14\), \(m_3=\frac{58 - 28}{3 - 2}=30\). Since the slope is not constant, it is not linear.
For an exponential function \(y = a\cdot b^x\). Using the point \((0,7)\), \(a = 7\) (since when \(x = 0\), \(y=a\cdot b^0=a\)).
Using the point \((1,14)\): \(14=7\cdot b^1\), so \(b = 2\). The equation is \(y=7\cdot2^x\)

Answer:

a. Linear, \(y = 3x-1\)
b. Exponential, \(y=7\cdot2^x\)