Looking at the possible answers, the only one that fits is D.
Let's first look at A: y = 2x
We know this doesn't work, since if you plug in 0 for x, y = 0, not 4.
Let's look at B: 2x^2
We know this doesn't work either, since if x is 0, y = 0, not 4.
Let's look at C: 2 × 2^x
We know this doesn't work, since if x = 0, y = 2:
![y = 2 * {2}^(x) \\ 4 = 2 * {2}^(0) \\ 4 = 2 * 1 \\ 4 = 2](https://img.qammunity.org/2019/formulas/mathematics/college/fv911txxhdnlaajbd3tii7v77zatnoks7a.png)
This leaves D: 4 × 2^x:
![y = 4 * {2}^(x) \\ 4 = 4 * {2}^(0) \\ 4 = 4 * 1 \\ 4 = 4](https://img.qammunity.org/2019/formulas/mathematics/college/9y5gzuwjngmlrx6tngwrxt9fp7tsadzr73.png)
It also works for (1,8) and (2,16):
![y = 4 * {2}^(x) \\ 8 = 4 * {2}^(1) \\ 8 = 4 * 2 \\ 8 = 8](https://img.qammunity.org/2019/formulas/mathematics/college/3dui6mmskfetnpf54cuyvh0guu46bfvdf5.png)
![y = 4 * {2}^(x) \\ 16 = 4 * {2}^(2) \\ 16 = 4 * 4 \\ 16 = 16](https://img.qammunity.org/2019/formulas/mathematics/college/e2udks86ereia6meqmlc36myxmd60l7qcu.png)
The answer is D.