Question

look at this table:

x	y
5	250
6	360
7	490
8	640

write a linear ($y = mx + b$), quadratic ($y = ax^2$), or exponential ($y = a(b)^x$) function that models the data.
$y = \square$

Explanation:

Step1: Check linear fit (constant slope)

Calculate slopes between points:
$\frac{250-160}{5-4}=90$, $\frac{360-250}{6-5}=110$, $\frac{490-360}{7-6}=130$, $\frac{640-490}{8-7}=150$
Slopes are not constant, so not linear.

Step2: Check quadratic fit (constant second difference)

First differences: $90, 110, 130, 150$
Second differences: $110-90=20$, $130-110=20$, $150-130=20$
Second differences are constant, so quadratic.

Step3: Solve for $a$ in $y=ax^2$

Use point $(4,160)$:
$160 = a(4)^2$
$160 = 16a$
$a = \frac{160}{16}=10$

Step4: Verify with another point

Use $(5,250)$: $y=10(5)^2=10*25=250$, which matches.
Use $(8,640)$: $y=10(8)^2=10*64=640$, which matches.

Answer:

$y=10x^2$