Graphing an exponential Function and asymptotes

[Relz]

Graph the curve y=[3(1/2)^x]−1
and determine the horizontal asymptote.

I have found the points(-5, 95) (-4, 47) (-3, 23) (-2, 11) (-1, 5) (0, 2) (1, 0.5) (2, -0.25) . I'm not sure how the graph is supposed to look or where my horizontal asymptote should be on the graph.

 

[mmm4444bot]

y=3(1

2)^x−1

We use the caret symbol ^ to text exponentiation.

I'm not sure how the graph is supposed to look.

After you plot the points and draw a smooth curve through them, you will know what the graph looks like.

However, if you'd like to have some idea of the general shape before plotting, then realize that the graph of y=(1/2)^x looks similiar to y=e^(-x), and subtracting 1 shifts the graph down one unit. The factor 3 alters the curve somewhat, too.

The asymptotic behavior occurs as x gets big, right? (Think about a general, decreasing exponential curve.)

So, I suggest that you calculate y for a few larger values of x, and the asymptote should become clear.

(5, ?) (10, ?) (15, ?) (20, ?)

Let us know, if you're still stuck.

 

[Relz]

Got it, thank you! I didn't even try the higher values before and that's where my asymptote was lying.