Anyone know if this is right?

f(x) = 2(4^x)

Step-by-step explanation:

Look at x = 0 in the table. It corresponds to f(x) = 2.

Now look at your

f(x) = 4^x

Use x = 0:

f(0) = 4^0 = 1, but the table has f(0) = 2, so your answer is not correct.

Notice the numbers in the f(x) column. They are 2, 8, 32, 128.

Notice that they are 2 times perfect squares.

2 = 2 * 1

8 = 2 * 4

32 = 2 * 16

128 = 2 * 64

Try

f(x) = 2(4^x)

Now you have

f(0) = 2(4^0) = 2 * 1 = 2

f(1) = 2(4^1) = 2 * 4 = 8

f(2) = 2(4^2) = 2 * 16 = 32

f(3) = 2(4^3) = 2 * 64 = 128

and all the given points work.

f(x) = 2(4^x)

check the picture below.

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.