Question

is this function linear, quadratic, or exponential?

x	y
-5	$\frac{320}{243}$
-4	$\frac{160}{81}$
-3	$\frac{80}{27}$
-2	$\frac{40}{9}$
-1	$\frac{20}{3}$

options:
linear
quadratic
exponential

Explanation:

Step1: Check linearity (constant Δy)

Calculate differences between consecutive y-values:
$\frac{160}{81}-\frac{320}{243}=\frac{480-320}{243}=\frac{160}{243}$
$\frac{80}{27}-\frac{160}{81}=\frac{240-160}{81}=\frac{80}{81}$
Differences are not constant, so not linear.

Step2: Check quadraticity (constant Δ²y)

Calculate differences of the differences:
$\frac{80}{81}-\frac{160}{243}=\frac{240-160}{243}=\frac{80}{243}$
$\frac{40}{9}-\frac{80}{27}=\frac{120-80}{27}=\frac{40}{27}$
Second differences are not constant, so not quadratic.

Step3: Check exponentiality (constant ratio)

Calculate ratios of consecutive y-values:
$\frac{\frac{160}{81}}{\frac{320}{243}}=\frac{160}{81}\times\frac{243}{320}=3$
$\frac{\frac{80}{27}}{\frac{160}{81}}=\frac{80}{27}\times\frac{81}{160}=3$
$\frac{\frac{40}{9}}{\frac{80}{27}}=\frac{40}{9}\times\frac{27}{80}=3$
$\frac{\frac{20}{3}}{\frac{40}{9}}=\frac{20}{3}\times\frac{9}{40}=3$
Ratios are constant (equal to 3).

Answer:

exponential