Answer:
[tex]\huge\boxed{\text{It's an exponential function}}\\\boxed{f(x)=16\left(\dfrac{1}{4}\right)^x}[/tex]
Step-by-step explanation:
[tex]\begin{array}{c|c|c|c|c|c}x&\ -3\ &-2\ &-1&\ 0&1\end{array}\\\overline{\begin{array}{c|c|c|c|c|c}y&1024&256&\ 64&16&4\end{array}}\\\\\text{It's not a linear function, because:}\\\\\text{in a linear function }\dfrac{y_2-y_1}{x_2-x_1}=constant\\\\\dfrac{-2-(-3)}{256-1024}=\dfrac{1}{-768}=-\dfrac{1}{786}\\\\\dfrac{-1-(-2)}{64-256}=\dfrac{1}{-192}=-\dfrac{1}{192}[/tex]
[tex]\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(\dfrac{1}{b}\right)^3=64\to\dfrac{1}{b}=\sqrt[3]{64}\\\\\dfrac{1}{b}=4\to b=\dfrac{1}{4}\\\\\boxed{f(x)=16\left(\dfrac{1}{4}\right)^x}[/tex]
[tex]\text{Check for other pairs:}\\\\(-2,\ 256)\\\\16\left(\dfrac{1}{4}\right)^{-2}=16(4)^2=(16)(16)=256\qquad \bold{CORRECT}\\\\(-1,\ 64)\\\\16\left(\dfrac{1}{4}\right)^{-1}=16(4)^1=(16)(4)=64\qquad\bold{CORRECT}\\\\(1,\ 4)\\\\16\left(\dfrac{1}{4}\right)^1=(16)\left(\dfrac{1}{4}\right)=\dfrac{16}{4}=4\qquad\bold{CORRECT}[/tex]
