2 Answers

Vanilla Face · Answer 1 · 2020-11-27T02:16:33+0000

Answer:

(1,12) is correct if you meant
$3 \cdot 4^x$ .

Please correct if I'm wrong about your expression.

Explanation:

I think you mean
$f(x)=3 \cdot 4^x$ .

Let's test the point and see.

A. (0,12)?

(0,12)=(x,y)

What happens when x equals 0? Is the result 12?

$3 \cdot 4^0$

$3 \cdot 1$

3(1)

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?

$3 \cdot 4^0$

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?

$3 \cdot 4^1$

$3 \cdot 4$

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?

$3 \cdot 4^(12)$

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
$3 \cdot 4^x$ .

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