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:
[tex]y = 2 \times {2}^{x} \\ 4 = 2 \times {2}^{0} \\ 4 = 2 \times 1 \\ 4 = 2[/tex]
This leaves D: 4 × 2^x:
[tex]y = 4 \times {2}^{x} \\ 4 = 4 \times {2}^{0} \\ 4 = 4 \times 1 \\ 4 = 4[/tex]
It also works for (1,8) and (2,16):
[tex]y = 4 \times {2}^{x} \\ 8 = 4 \times {2}^{1} \\ 8 = 4 \times 2 \\ 8 = 8[/tex]
[tex]y = 4 \times {2}^{x} \\ 16 = 4 \times {2}^{2} \\ 16 = 4 \times 4 \\ 16 = 16[/tex]
The answer is D.
