Respuesta :
Answer:
(1,12) is correct if you meant [tex]3 \cdot 4^x[/tex].
Please correct if I'm wrong about your expression.
Step-by-step explanation:
I think you mean [tex]f(x)=3 \cdot 4^x[/tex].
Let's test the point and see.
A. (0,12)?
(0,12)=(x,y)
What happens when x equals 0? Is the result 12?
[tex]3 \cdot 4^0[/tex]
[tex]3 \cdot 1[/tex]
[tex]3(1)[/tex]
[tex]3[/tex]
Yep that isn't 12 so (0,12) is not on the graph of f.
B. (0,0)?
(0,0)=(x,y)
What happens when x equals 0? Is the result 0?
[tex]3 \cdot 4^0[/tex]
We already this and got 3 so (0,0) is not on the graph of f.
C. (1,12)?
(1,12)=(1,12)
What happens when x equals 1? Is the result 12?
[tex]3 \cdot 4^1[/tex]
[tex]3 \cdot 4[/tex]
[tex]12[/tex]
The result is 12 so (1,12) is on the graph of f.
C. (12,1)
(12,1)=(x,y)
What happens when x equals 12? Is the result 1?
[tex]3 \cdot 4^{12}[/tex]
This will result in a really big number that isn't 1 so (12,1) is not on the graph of f.
(1,12) is correct if you meant [tex]3 \cdot 4^x[/tex].
