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

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Answer 1 · 2015-08-06T01:17:38

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

$y=40^x$

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.

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