Analyzing graphs of exponential functions: negative initial value | High School Math | Khan Academy

So we have a graph here of the function ( f(x) ) and I'm telling you right now that ( f(x) ) is going to be an exponential function. It looks like one, but it's even nicer. When someone tells you that, and our goal in this video is to figure out at what ( x ) value—so when—when does ( f(x) ) equal -125?

You might be tempted to just eyeball it over here, but when ( f(x) ) is -125, that's like right below the x-axis. So if I tried to eyeball it, it would be very difficult. It's very difficult to tell what value that is; it might be at 3, it might be at 4, I am not sure. So instead of actually—well, maybe I don't want to just eyeball it and guess it—instead, I'm going to actually find an expression that defines ( f(x) ) because they've given us some information here, and then I can just solve for ( x ).

So, let's do that. Well, since we know that ( f(x) ) is an exponential function, we know it's going to take the form ( f(x) = a \cdot r^x ). Well, the initial value is straightforward enough; that's going to be the value that the function takes on when ( x ) is equal to 0. You could even see here if ( x = 0 ), the ( r^x ) would just be 1, and so ( f(0) ) will just be equal to ( a ).

And so what is ( f(0) )? Well, when ( x = 0 ), this essentially—we're saying where does it intersect? Where does it intersect the y-axis? We see ( f(0) = -25 ), so ( a = -25 ). When ( x ) is 0, the ( r^x ) is just 1, so ( f(0) ) is going to be -25; we see that right over there.

Now to figure out the common ratio, there are a couple of ways you could think about it. The common ratio is the ratio between two successive values that are separated by one. What do I mean by that? Well, you could view it as the ratio between ( f(1) ) and ( f(0) ); that's going to be the common ratio, or the ratio between ( f(2) ) and ( f(1) )—that is going to be the common ratio.

Well, lucky for us, we know ( f(0) = -25 ), and we know that ( f(1) = -5 ). So just like that, we're able to figure out that our common ratio ( r ) is ( -5 / -25 ), which is the same thing as ( 1/5 ). Divide a negative by negative; you get a positive. So you're going ( 5 / 25 ), which is ( 1/5 ).

So now we can write an expression that defines ( f(x) ). ( f(x) ) is going to be equal to ( -25 \cdot (1/5)^x ). And so let's go back to our question: When is this going to be equal to -125?

So when does this equal -125? Well, let's just set them equal to each other. So let—there's a siren outside, I don't know if you hear it—so negative; I'll power through. Alright, negative. So let's see, at what ( x ) value does this expression equal -125?

Let's see, we can multiply—well, actually we want to solve for ( x ). So let's see, let's divide both sides by -25 and so we are going to get ( (1/5)^x = (-125) / (-25) ). This -25 is going to go away, and on the right-hand side, we're going to have—dividing negative by negative, it's going to be positive—it's going to be ( 1 / 5 ).

And ( (1/5)^x ) is the same thing as ( 1^x / 5^x ) is equal to ( 1 / 5 ). So we can see that ( 5^x ) needs to be equal to 625.

So let me write that over here. ( 5^x = 625 ). Now, the best way I could think of doing this is let's just think about our powers of 5. So ( 5^1 = 5 ), ( 5^2 = 25 ), ( 5^3 = 125 ), ( 5^4 = 625 ). So ( x ) is going to be 4, because ( 5^4 = 625 ).

So we can now say that ( f(4) ) is equal to -125. Once again, you can verify that; you can verify that right over here: ( (1/5)^4 = 1 / 625 ). ( -25 / 625 ) is going to be -125.

So hopefully that clears things up a little bit.