How do you graph #y= 40^x #?

1 Answer
Aug 6, 2015

Answer:

Find the intercepts and asymptotes, plot a few points, and then sketch the graph.

Explanation:

#y=40^x#

Step 1. Find the domain and range.

#y# is defined for all real values of #x#, so the domain is the set of all real numbers.

#40^x>0#, so the range is #y>0#

Step 2. Find the #y#-intercept.

Let #x=0#.

#y= 40^0 = 1#

The #y#-intercept is at (#0,1#).

Step 3. Find the #x#-intercept.

There is no #x#-intercept, because the range is #y>0#.

Step 4. Find the horizontal asymptote.

As #x# decreases without bound, #40^x# approaches #0#.

The horizontal asymptote is at #y=0#.

Step 5. Calculate some extra points.

We have one point: #f(0) =1#.

#f(1/2) =40^(1/2) = sqrt40 ≈ 6.3#

#f(-1/2) = 40^(-1/2) = 1/sqrt40 ≈0.16#

Step 6. Plot the axes and your three points.

Graph 1

Step 7. Complete the graph with a smooth curve through the three points.

Graph 2

And you have your graph.