Answer:
1(2)^{x}
Step-by-step explanation:
The easiest way to approach this problem in my opinion is to take each option in turn and work out if it holds true for all x.
So, taking them one by one:
a) 2x+1
When x = 0, y = 2(0)+1 = 0+1 = 1
When x = 1, y = 2(1) + 1 = 2+1 = 3 -- This is wrong as when x=1 y=2 from the table so it can't be this one.
b) 1(2)^{x} -- We can rewrite this for simplicity as just 2^{x} as anything times 1 is itself.
So, x=0 y=2^{0}=1 (anything to the power of 0 = 1)
x=1 y=2^{1}=2
x=2 y=2^{2}=4
x=3 y=2^{3}=8
x=4 y=2^{4} = 16
This is correct so you can stop here and select this answer.
For extra confidence you can find examples for x that disprove the next 2:
c) y=x+2
When x = 3, y would be 3+2 = 5, which is wrong as it should be 8
d) 2(1)^{x}
This is just 2*1, we know from bidmas/pemdas that you do powers first, so 1^{x} is always 1, and then you multiply by 2, so this is 2 for all values of x, which is again wrong.
Any questions let me know :)
