Question

Which is the graph of f(x) = (2)-x ?
On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and increases into quadrant 1. It crosses the y-axis at (0, 1).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases into quadrant 2. It crosses the y-axis at (0, 1).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases into quadrant 2. It crosses the y-axis at (0, 2).

Answer 1

Answer:

We will choose option 2.

Step-by-step explanation:

The given function is $f(x) = 2^{-x} = \frac{1}{2^{x} }$

So, from the above equation it is clear that as x increases in the first quadrant the value of f(x) tends to zero and increases into second quadrant.{ Since x is negative in the second quadrant}

Again, for x = 0, y becomes 1, i.e. the function crosses the y axis at (0,1).

Therefore, on a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases into quadrant 2. (0, 1) point is the point of intersection with the y-axis.  

So, we will choose option 2. (Answer)

Answer 2

Answer:

The correct option is B) On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases into quadrant 2. It crosses the y-axis at (0, 1).

Step-by-step explanation:

Consider the provided equation.

$f(x) = (2)^{-x}$

Draw the graph of the equation by using the table shown below:

x      f(x)

-2      4

-1       2

0       1

1     0.5

2    0.25

Now draw the graph by using the table.

By observe the graph the exponential function approaches y = 0 in quadrant 1.

The function increases into quadrant 2 and It crosses the y-axis at (0, 1).

Hence, the correct option is B) On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases into quadrant 2. It crosses the y-axis at (0, 1).