Question

look at this table:

x	y
2	-20
3	-40
4	-80
5	-160

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 linearity (constant slope)

Slope between (1,-10) & (2,-20): $\frac{-20 - (-10)}{2-1} = -10$
Slope between (2,-20) & (3,-40): $\frac{-40 - (-20)}{3-2} = -20$
Slopes are not equal, so not linear.

Step2: Check quadratic (constant 2nd difference)

1st differences: $-20-(-10)=-10$, $-40-(-20)=-20$, $-80-(-40)=-40$, $-160-(-80)=-80$
2nd differences: $-20-(-10)=-10$, $-40-(-20)=-20$, $-80-(-40)=-40$
2nd differences are not equal, so not quadratic.

Step3: Check exponential (constant ratio)

Ratio of consecutive $y$-values: $\frac{-20}{-10}=2$, $\frac{-40}{-20}=2$, $\frac{-80}{-40}=2$, $\frac{-160}{-80}=2$
Ratio $b=2$ is constant, so exponential.

Step4: Solve for $a$ in $y=a(b)^x$

Use $x=1, y=-10$: $-10 = a(2)^1$
Solve for $a$: $a = \frac{-10}{2} = -5$

Step5: Write the final function

Substitute $a=-5, b=2$ into $y=a(b)^x$

Answer:

$y = -5(2)^x$