Current time:0:00Total duration:7:38

0 energy points

# Writing exponential functions from graphs

Video transcript

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 0 pass
through the points negative 1 comma 9-- so this
is negative 1 comma 9 right over here--
and 1 comma 1. 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 r is greater than
0, but it's a pretty good hint that r is going to
be between 0 and 1. The fact that g of
x keeps approaching, it's getting closer and closer
and closer to 0 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, right
over here, that's our slope, that's our change
in y over change in x. The rate of 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,
if we were going from x equals negative
1 to x equals 1. So we could think of it
as, we're finishing at 1, we started at negative 1. So 1 minus negative 1,
our change in x is 2. We see that right over there. And what about our change in y? Well, we start at 9. Let me do this in
maybe another color here just-- We start at
9, and we end up at 1. So we end up at 1. We started at 9. 1 Minus 9 is negative 8. And just to be clear,
when x is 1, y is 1. When x is negative 1, y is 9. Another way to think about it,
the way I drew it right over here, we're finishing at
x equals 1, y equals 1. We started at x equals
negative 1, y equals 9. And so we just took the
differences, we get negative 8 over 2, which is
equal to negative 4. And so now we can
write that f of x is equal to negative 4,
that's our slope, times x. Negative 4 times x plus b. And you can see that
slope right over here. Every time you
increase your x by 1-- you got to be careful here--
but every time you increase your x by 1, you're
decreasing your y. And here on the x-axis,
we're marking off every 1/2. So every time you
increase your x by 1, you are decreasing
your y by 4 there. So that makes sense that
the slope is negative 4. 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 b. So let's try f of,
let's try 1, because 1 is a nice simple number. So we could write
f of 1, which would be negative 4 times 1 plus b. And they tell us that f
of 1 is 1, is equal to 1. And so this part
right over here, we could write that as negative
4 plus b is equal to 1. And then we could add 4 to
both sides of this equation, and then we get b is equal to 5. So we get f of x is equal
to negative 4 x plus 5. And now does that make sense,
that the y-intercept here is 5? Well, you see that right
over here by inspection, you could have guessed actually
that the y-intercept here is 5. now we've solved it. Maybe this was
5.00001 or something, but now we know for sure. It's negative 4 x plus 5. Or another way you
could have said it, if the slope is negative 4,
if this right over here is 9, you increase 1 in
the x direction, you're going to decrease
4 in the y direction. That will get you
to y is equal to 5. 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
just try this first point. So g of negative 1, which if we
look at this right over here, would be a times r
to the negative 1. They tell us that g of negative
1 is going to be equal to 9. And so, we could write this
a times r to the negative 1, that's the same thing as
a over r is equal to 9. Or we can multiply
both sides by r. And we could say
a is equal to 9r. Now let's use this other point. They tell us that
g of 1, which would be the same thing as a
times r to the first power, or just a times r, that
that is equal to 1. Or a times r is equal to 1. So how can we use
this information right here, a is equal to 9r
and a times r is equal to 1, 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 in a here, I'll write 9r times
r is equal to 1. Or we could write--
let me scroll down a little bit-- we could write
9r squared is equal to 1. Divide both sides by 9. r squared is equal to 1 over 9. And now to figure
out r, you might want to take the positive
and negative square root of both sides. But they tell us that
r is greater than 0. 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 9 times r. So 9 times 1/3, a is equal to 3. So our exponential
function could be written as g of x is equal
to a, which is 3, times r, which is 1/3 to the 1/3,
to the x power. Or you could even
write it like this because order of
operations would imply that you do the
exponential first. Watch, I'll write
it like this again, just to make it clear this
isn't just 1 to the third power, this is 1/3 to the third power.