Question

X= -3, -2, -1, 0, 1 y= 1024, 256, 64, 16, 4 find either an exponential or linear function

Answer 1

`\huge\boxed{\text{It's an exponential function}}\\\boxed{f(x)=16\left((1)/(4)\right)^x}`

Explanation:

`\begin{array}cx&\ -3\ &-2\ &-1&\ 0&1\end{array}\\\overline{\begin{array}cy&1024&256&\ 64&16&4\end{array}}\\\\\text{It's not a linear function, because:}\\\\\text{in a linear function }(y_2-y_1)/(x_2-x_1)=constant\\\\(-2-(-3))/(256-1024)=(1)/(-768)=-(1)/(786)\\\\(-1-(-2))/(64-256)=(1)/(-192)=-(1)/(192)`

`\text{Therefore it's an exponential function}\ f(x)=ab^x\\\\\text{For}\ (0,\ 16)\ \text{we have}\\\\f(0)=16\to ab^0=16\to a=16\\\\\text{For}\ (-3,\ 1024)\ \text{we have}\\\\f(-3)=1024\to 16b^(-3)=1024\qquad\text{divide both sides by 16}\\\\b^(-3)=64\\\\\left((1)/(b)\right)^3=64\to(1)/(b)=\sqrt[3]{64}\\\\(1)/(b)=4\to b=(1)/(4)\\\\\boxed{f(x)=16\left((1)/(4)\right)^x}`

`\text{Check for other pairs:}\\\\(-2,\ 256)\\\\16\left((1)/(4)\right)^(-2)=16(4)^2=(16)(16)=256\qquad \bold{CORRECT}\\\\(-1,\ 64)\\\\16\left((1)/(4)\right)^(-1)=16(4)^1=(16)(4)=64\qquad\bold{CORRECT}\\\\(1,\ 4)\\\\16\left((1)/(4)\right)^1=(16)\left((1)/(4)\right)=(16)/(4)=4\qquad\bold{CORRECT}`