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### Course: Algebra 1 > Unit 12

Lesson 6: Exponential functions from tables & graphs- Writing exponential functions
- Writing exponential functions from tables
- Exponential functions from tables & graphs
- Writing exponential functions from graphs
- Analyzing tables of exponential functions
- Analyzing graphs of exponential functions
- Analyzing graphs of exponential functions: negative initial value
- Modeling with basic exponential functions word problem
- Connecting exponential graphs with contexts

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# Writing exponential functions from graphs

Given a graph of a line, we can write a linear function in the form y=mx+b by identifying the slope (m) and y-intercept (b) in the graph.
GIven a graph of an exponential curve, we can write an exponential function in the form y=ab^x by identifying the common ratio (b) and y-intercept (a) in the graph. Created by Sal Khan.

## Want to join the conversation?

- general form of y= r . a^x is never really introduced--it just sort of appears in the context of problem videos. i found this confusing as i had not seen this before(25 votes)
- Actually, the general form of an exponential function isn't f(x)= a*r^x, which is why I believe that it is not formally introduced in these videos. Really, the question just wants you to format your answer that way. And to be quite honest, f(x)=a*r^x is the most logical way to format your answer. The general equation for an exponential function is f(x)=ab^(x-h) +k not f(x)=a*r^x. I hope that answers your question!(25 votes)

- At3:38, couldn't he just find the y-intercept using the graph?(13 votes)
- Yes he can, in this case it would have been very straightforward. for f(x), (0,5) and for g(x), (0,3). Then you can use point (1,1) (or the other point, but (1,1) is simpler) to find m and r respectively.

I think the intent was to show another way of doing it, as what you suggested was shown in a previous video(12 votes)

- At3:31couldn't Sal have just looked at the graph of f(x) to see that the y-intercept (b) was 5? This won't be true in every case, but when the graph passes directly through a number that is marked on the graph you can read the graph at that number. Right?(6 votes)
- In this case, you may be allowed to read the graph because the equation is linear. It is a perfectly straight line, and since (0,5) is exactly between (-1,9) and (1,1) it is easy to tell that that is the y-intercept. But yes, you should always check it, if only so your teacher doesn't get mad. :)(8 votes)

- I'm totally lost from about5:25on. How does ar to the -1 become a/r?(5 votes)
- How did r^2=1/9 become r=1/3 ?(4 votes)
- When you square root one side of the equation, you square root the other side of the equation. In the video, he took the square root of 1/9. The square root of 1 is 1, and the square root of 9 is 3, so the square root of 1/9 is 1/3. Does this make sense?(4 votes)

- why does sal always calculate the slope first? is there a reason for it (like, maybe it's necessary to do it that way in later math), or is it just a habit? because i usually look at the y intercept, then plug in a random set of points on the line to find the slope that way. i always mess up (proper) slope calculations for some reason, so doing it this way has a higher success rate with me, but if there's a reason to favor this method over my own then i'd like to know so i can get used to it sooner.(2 votes)
- In these practice exercises we conveniently have integer for the y-intercepts so it is easy to find them on the graph, but in real life sets of data, the line will cross the y-axis in-between 2 integers and you won't be able to do that :-)

You'll have to find two points that you are sure of, then calculate the slope first.

Good luck in your practice!(6 votes)

- I have a question. I looked at this video and learned how to find the equation of exponential functions, but how would you find an exponential function if it had dilations, vertical translations, and horizontal translations?(2 votes)
- All functions basically react the same to dilations and translations. The most basic exponential function is y=2^x. Looking at some of the key features, you have no x intercept, a y intercept at 1 (when x=0, 2^0=1) (0,1), then additional points of (1,2)(2,4)(3,8)(4,16)(-1,1/2)(-2,-1/4) etc. The horizontal asymptote is at y=0. So you could start adding changes to this such as y=a 2^(x-b)+c. If a is negative, it reflects across x. If a is >1, it multiplies all x values by a (for example if a=2, you would get (-2,1/2)(-1,1)(0,2)(1,4)(2,8). If a<1, then it cuts the points by the scale factor. If b is negative, it shifts to right, and if b is positive it shifts left, if c is negative it moves down and if c is positive, it moves up (the asymptote is moved up or down c units).(3 votes)

- Will this same process work for exponential growth?(3 votes)
- Yes, it should(1 vote)

- At6:48, why is Sal able to just take the principal root? Shouldn't there be any mathematical expresion that can let me legitimally rule r= -1/3 out?(2 votes)
- Well, for a graph like that, you could just look at when
`x = 0`

, it is 3. So`a = 3`

. Then you just pick a point for example, that point (1, 1). You will now have`1 = 3 * r^1`

which is the same as`1 = 3r`

. You can divide both sides by 3 which leaves you with`1/3 = r`

saving quite a lot of space.

But if you just have two points, you'll need to figure out the equation yourself.(2 votes)

- 6:00what makes the system non linear?(1 vote)
- The r^2 makes it non-linear. A linear system would not have an variable with an exponent of 2. All variables would have an exponent of 1.(3 votes)

## Video transcript

- [Instructor] The graphs of
the linear function f of x is equal to mx plus b and the exponential function g of x is equal to a times r to the x where r is greater than zero pass through the points
negative one comma nine, so this is negative one
comma nine right over here, and one comma one. Both graphs are given below. So this very clearly
is the linear function. It is a line right over here. And this right over here is
the exponential function, so right over here. And given the fact that
this exponential function keeps decreasing as x gets
larger and larger and larger is a pretty good hint that
our r right over here, they tell us that r is greater than zero, but it's a pretty good hint that r is going to be
between zero and one. The fact that g of x keeps approaching, it's getting closer and
closer and closer to zero as x increases. But let's use the data they're giving us, the two points of intersection, to figure out what the equations
of these two functions are. So first we can tackle
the linear function. So f of x is equal to mx plus b. So they give us two points. We could use those points
first to figure out the slope. So our m, our m right over
here, that's our slope. That's our change in y over change in x. The rate and change of the vertical axis with respect to the horizontal axis. So let's see, between those two points, what is our change in x? Our change in x, we're going from x equals
negative one to x equals one. X equals one. So we could think of it
as we're finishing at one. We started at negative one. So one minus negative one,
our change in x is two. We see that right over there. And what about our change in y? Well, we start at nine. Let me do this in maybe
another color here. We start at nine, and we end up at one. So we end up at one. We started at nine. One minus nine is negative eight. And just to be clear,
when x is one, y is one. When x is negative one, y is nine. Another way to think about it, the way I drew it right over here, we're finishing at x
equals one, y equals one. We started at x equals
negative one, y equals nine, and so we just took the differences. We get negative eight over two, which is equal to negative four. Which is equal to negative four. And so now we can write that f of x is equal to negative four, negative four, that's our slope, times x. Negative four times x plus b. And you can see that
slope right over here. Every time you increase your... Every time you increase your x by one. I gotta be careful here,
I got a little bit. Every time you increase your x by one, you're decreasing your y. And here on the x axis,
we're marking off every half. So every time you increase your x by one, you are decreasing your y. You are decreasing your y by four there, so that makes sense that
the slope is negative four. So now let's think about what b is. So to figure out b, we could
use either one of these points to figure out, given an x, what f of x is, and then we can solve for by. Let's try one, 'cause one
is a nice, simple number. So we could write f of one, which would be negative
four times one plus b. And they tell us that f of
one is one, is equal to one. And so this part right over here, we could write that as negative
four plus b is equal to one, and then we could add four to
both sides of this equation, and then we get b is equal to five. So we get f of x is equal
to negative 4x plus five. Now, does that make sense that the y-intercept here is five? Well, you see that right over here. By inspection, you could
have guessed, actually, that the y-intercept here is five, but now we've solved it. Maybe this was 5.00001 or something, but now we know for sure
it's negative 4x plus five. Or another way you could've said it, if the slope is negative four, if this right over here is nine, you increase one in the x direction, you're gonna decrease
four in the y direction, and that will get you
to y is equal to five, so that is the y-intercept. But either way, we have figured
out the linear function. Now let's figure out the
exponential function. So here we could just use the two points to figure out these two unknowns. So, for example, let's
try this first point. So g of negative one, which if we look at this right over here, would be a times r to the negative one. They tell us that g of negative one is going to be equal to nine. G of negative one is equal to nine. And so we could write this a
times r to the negative one. That's the same thing as
a over r is equal to nine, or we could multiply both sides by r, and we could say a is equal to 9r. Now let's use this other point. This other point, they tell us. They tell us that g of one, which would be the same thing as a times r to the first
power or just a times r, that that is equal to one,
or a times r is equal to one. So how can we use this
information right here, a is equal to 9r and a
times r is equal to one, to solve for a and r? Well, I have a little system here. It is a non-linear system,
but it's a pretty simple one. We could just take this a and substitute it in
right over here for a, and so we would get 9r for a. This first constraint tells
us a must be equal to 9r. So 9r, instead of writing
an a here, I'll write 9r, times r is equal to one. Or we could write, let me
scroll down a little bit. We could write 9r squared is equal to one. Divide both sides by nine. R squared is equal to one over nine. And now to figure out r, you might wanna take the
positive and negative square root of both sides, but they tell
us that r is greater than zero, so we can just take the
principal root of both sides and we get r is equal to 1/3. And then we could substitute this back into either one of these other
two to figure out what a is. We know that a is equal to nine times r. So nine times 1/3, a is equal to three. So our exponential function
could be written as g of x is equal to a, which is
three, times r, which is 1/3, 1/3 to the x power.