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this is from the graph basic exponential functions on Khan Academy and they ask us graph the following exponential function and they give us the function H of X is equal to 27 times 1/3 to the X so our initial value is 27 and 1/3 is our common ratio it's written in kind of standard exponential form and they give us this little graphing tool where we can define these two points and we can also define we can define a horizontal asymptote to construct our function and these three things are enough to define - to graph and exponential if we know that it is an exponential function so let's think about it a little bit so the easiest thing that I could think of is well let's think about its initial value its initial value is going to be when x equals 0 x equals 0 it's 1 3 to the 0 power which is just 1 and so you're just left with 27 times 1 or just 27 that's why we call this number here when you're written it in this form you call this the initial value so when X is equal to 0 H of X is equal to 27 and we're graphing y equals H of X so now let's graph another point so let's think about it a little bit when when X is equal to 1 when X is equal to 1 what is H of X it's going to be one-third to the first power which is just 1/3 and so 1/3 times 27 is going to be 9 so when H when X is 1 H of 1 is 9 and we can verify well and now let's just think about let's think about the asymptotes so what's going to happen here when X becomes really really really really really really big well if I take 1/3 to like a really large exponent to say to the to the 10th power to the hundredth power or to the thousandth power this thing right over here is going to start approaching zero as we get as X becomes much much much much larger and so something that is approaching 0 times 27 well that's going to approach 0 as well so we're going to have a horizontal asymptote at 0 and you can verify that this works for more than just the two points we thought about when X is equal to 2 this is telling us that the graph y equals H of X goes through the point 2 comma 3 so H of 2 should be equal to 3 and you can verify that that is indeed the case if X is to 1 third squared is 9 oh sorry one third squared is 1/9 times 27 is 3 and we see that right over here when X is 2 H of 2 is 3 so I feel pretty good about that let's do another one of these so graph the following exponential function so same logic X is when X is zero the G of zero is just going to it's just going to boil down to that initial value and so let me scroll down the initial value is negative 30 and so let's think about when X is equal to 1 when X is equal to 1 2 to the first power is just 2 and so 2 times negative 30 is negative 60 so when X is equal when X is equal to 1 the value of the graph is negative 60 now let's think about let's think about this asymptote where that should sit so let's think about what happens when X becomes really really really really really really negative so when X is really negative so 2 to the negative 1 power is 1/2 to the negative 2 is 1/4 to the negative 3 is 1/8 as you get larger and larger negative at or higher magnitude negative values or in other ways says X becomes more and more and more negative 2 to that power is going to approach 0 and so negative 30 times something approaching 0 is going to approach zero so this asymptotes in the right place our horizontal asymptote as X as X approaches negative infinity as we move further and further to the left the value of the function is going to approach zero and we can see it kind of approaches zero from below we can see that it approaches below because we already just looked at the initial value and we use that common ratio to find one other point hopefully you found that interesting