Question

21.

x	0	1	3	4
y	5	10	20	40

model (circle one):
linear growth quadratic (local max) exponential growth
linear decay quadratic (local min) exponential decay
identify whether the function is linear, quadratic, or exponential?
options:
exponential growth
quadratic (local max)
linear decay
linear growth
exponential decay

Explanation:

Step1: Check linearity

For a linear function \(y = mx + b\), the slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\) should be constant.
For \(x = 0,y = 5\) and \(x = 1,y = 10\), \(m_1=\frac{10 - 5}{1-0}=5\)
For \(x = 1,y = 10\) and \(x = 3,y = 20\), \(m_2=\frac{20 - 10}{3 - 1}=5\)
For \(x = 3,y = 20\) and \(x = 4,y = 40\), \(m_3=\frac{40 - 20}{4 - 3}=20\)
Since the slopes are not constant, it is not a linear function.

Step2: Check quadratic form

A quadratic function is of the form \(y=ax^{2}+bx + c\). The second - difference for a quadratic function is constant.
First differences: \(10 - 5=5\), \(20 - 10 = 10\), \(40 - 20=20\)
Second differences: \(10 - 5 = 5\), \(20 - 10=10\)
Since the second differences are not constant, it is not a quadratic function.

Step3: Check exponential form

An exponential function is of the form \(y = a\cdot b^{x}\).
When \(x = 0,y=a\cdot b^{0}=a = 5\)
When \(x = 1,y=5\cdot b^{1}=10\), then \(b = 2\)
Check for \(x = 3\): \(y=5\cdot2^{3}=5\times8 = 40\) (but the table has \(y = 20\) when \(x = 3\), wrong. Let's check another approach.
We can also check the ratio of \(y\) - values. \(\frac{y_1}{y_0}=\frac{10}{5}=2\), \(\frac{y_2}{y_1}=\frac{20}{10} = 2\), \(\frac{y_3}{y_2}=\frac{40}{20}=2\)
The general form of an exponential function is \(y=y_0\cdot r^{x}\), where \(y_0 = 5\) (when \(x = 0\)) and \(r = 2\) (common ratio). So the function is \(y = 5\cdot2^{x}\), which is an exponential growth function (\(r>1\))

Answer:

Exponential Growth