Respuesta :

AL2006

You need to pick a point on the graph, then go through the list of choices
and see which one can produce the point you picked.

Look at the point on the graph right above the origin, where the graph
crosses the y-axis.  At that point, x=0 and f(x) (or 'y') = 1 .

Now look at the choices, one at a time, and see whether  (0, 1)  fits.
I'll start at the bottom of the list:

-->  f(x) = 4^(x-3)
When x=0, f(x) = 4^(-3) .
4^(-3) is not '1', so our point is not on this f(x).

-->  f(x) = 4^x + 3
When x=0, f(x) = 4⁰ + 3  =  1 + 3  =  4
'4' is not '1', so our point is not on this f(x).

-->  f(x) = 4^x - 3
When x=0, f(x) = 4⁰ - 3  =  1 - 3  =  -2
'-2' is not '1', so our point is not on this f(x).

(Only one choice left.)

-->  f(x) = 4^x
When x=0, 4^x = 4⁰  =  1
That's it !  Our point is on this f(x).
Which is lucky, because we were out of choices.


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