Graphing exponential functions | Mathematics III | High School Math | Khan Academy

We're told to use the interactive graph below to sketch a graph of ( y = -2 \cdot 3^x + 5 ).

And so this is clearly an exponential function right over here. Let's think about the behavior as ( x ) changes. When ( x ) is very negative or when ( x ) is very positive. When ( x ) is very negative, ( 3 ) to a very negative number—like you said, let's say you had ( 3^{-3} )—that would be ( \frac{1}{27} ), or ( 3^{-4} )—that'd be ( \frac{1}{81} ). So this is going to get smaller and smaller and smaller. It's going to approach ( 0 ) as ( x ) becomes more negative.

And since this is approaching ( 0 ), this whole thing right over here is going to approach ( 0 ). So this whole expression, if this first part's approaching ( 0 ), then this whole expression is going to approach ( 5 ). We're going to have a horizontal asymptote that we're going to approach as we go to the left. As ( x ) gets more and more negative, we're going to approach positive ( 5 ).

Then, as ( x ) gets larger and larger and larger, ( 3^x ) is growing exponentially. But then we're multiplying it times ( -2 ), so it's going to become more and more and more negative, and then we add a ( 5 ).

What we have here, well, this doesn't look like a line; we want to graph an exponential. So let's go pick the exponential in terms of ( x ). There you have it! We can move three things: we can move this point; it doesn't even just have to be the ( y )-intercept, although that's a convenient thing to figure out.

We can move this point here, and we can move the asymptote. Maybe the asymptote's the first interesting thing we said: as ( x ) becomes more and more and more and more negative, ( y ) is going to approach ( 5 ). So let me put this up here; that's our asymptote.

It doesn't look like it quite yet, but when we try out some values for ( x ) and the corresponding ( y ) values and we move these points accordingly, hopefully our exponential is going to look right.

So let's think about—let's pick some convenient ( x ) values. Let's think about when ( x = 0 ). If ( x = 0 ), ( 3^0 = 1 ); ( -2 \cdot 1 = -2); and ( -2 + 5 = 3 ). So when ( x = 0 ), ( y = 3 ).

Now, let's think about when ( x = 1 ). I’m just picking that because it's easy to compute: ( 3^1 = 3 ); ( -2 \cdot 3 = -6); and ( -6 + 5 = -1 ). So when ( x = 1 ), ( y = -1 ).

Let's see, is this consistent with what we just described? When ( x ) is very negative, we should be approaching positive ( 5 ), and that looks like the case. As we move to the left, we're getting closer and closer and closer to ( 5 ).

In fact, it looks like they overlap, but really we're just getting closer and closer and closer because this term right over here is getting smaller and smaller and smaller as ( x ) becomes more and more and more negative.

But then, as ( x ) becomes more and more positive, this term becomes really negative because we're multiplying it times ( -2 ), and we see that it becomes really negative.

So I feel pretty good about what we've just graphed. We've graphed the horizontal asymptote, it makes sense, and we've picked two points that sit on this graph of this exponential. So I can check my answer, and we got it right!