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Algebra 1
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 tables
Both linear equations and exponential equations represent relationships between two variables. However, the way that the variables are related to each other in each type of equation is different.
A linear equation can be thought of as representing repeated addition on an initial value, while an exponential equation can be thought of as representing repeated multiplication on an initial value. Created by Sal Khan.
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i didnt understand any of this(27 votes)
If you don't get it, check the hints in the exercise. If you don't get it here are some examples.
x | f(x)
---------
0 | 4
1 | 8
So, we see that when x is 0, f(x) = 4. f(x) is like y. Then, when x=1, f(x)[or y] = 8. We see that we have to multiply 4 by 2 to get 8. So, f(x)=4*(2)^x. OR, y=4*(2)^x. And when we plug in x=1, We get f(x)=8, which is right based on the table. I hope this helps. If it doesn't, just comment.(3 votes)
For g(x)= a*r^x, Sal immediately used g(0) to find out what a was, but what are you supposed to do if you are not told what g(o) is equal to? Would you still be able to solve the problem and if so how?(10 votes)
If you didn't know g(0), you could instead start with another point like (1, g(1)) and then work backwards from there to find g(0).
For instance, to get from g(0) to g(1), we know that we needed to multiply by 2/3. So, to go backwards, we would need to divide by 2/3, which is the same thing as multiplying by 3/2. So we could just take g(1), which is 2, and multiply that by 3/2 to get back to g(0). Since 2 * 3/2 = 3, we know that g(0) must be 3, and now we're done.
So, using other points, you can always work backwards to find g(0). I hope this helps.(7 votes)
I dont really understand this pls help :((5 votes)
So Sal talks about two functions, linear (with a common difference) and exponential (with a common ratio). It is much easier when x is incremented by 1 such as shown (1, 2, 3, ...) because it is easier to find what we need. For linear equations, we have y = m (slope) x + b (y intercept) and for exponential equations we have y = a (initial value)*r(ratio or base)^x. So in each case, we need to find two things.
In both cases, the y intercept and initial value are found where x = 0 (y intercept) and the table gives us these, so linear b = 5 and exponential a = 3. We are already 1/2 way there.
Linear slope is found by the common difference (since slope is change in y/change in x and change in x is 1, divide by 1 does not change anything). We subtract any two consecutive terms 2nd - 1st which gives 7 -5 = 9 - 7 = 11 - 9 = ... = 2. So with a slope of 2 and an intercept of 5, we get y = 2x + 5.
For exponential, we divide instead of subtract, so 2/3 = (4/3)/2 = (8/9)/(4/3) = ... = 2/3. We have to use our dividing fraction skills to see they are the same, flip denominator and multiply, so 8/9 * 3/4 does reduce to 2/3 which gives us the base. So with an initial value of 3 and a base (common ratio) of 2/3 we get y = 3 (2/3)^x. I hope running them together makes more sense than separate since they can be related to each other is a variety of ways. Does this help?(2 votes)
if Sal's got short memory then I dunno bout the rest of us(5 votes)
What do you do if you only have 2 points?
Here was the problem:
Complete the equation for the table:
h(x)=____
x___|0_|1
h(x)|10|4
I have h(x)=10*b^x. How do I find the exponential degree (b) if there are only two points? Thanks!(3 votes)
Hi Kaitlyn,
You are almost there.
You already determined what happens if x = 0:a·b⁰=10 <=> a·1=10 <=> a=10
Now do the same for x = 1, replacing "a" with 10:10·b¹=4 <=> 10b=4 <=> b=4/10 <=> b=2/5=> h(x)=10·(2/5)ˣ(2 votes)
- why does he do triangles when he says change for example at1:30(0 votes)
Those are not triangles, they are the Greek capital letter delta, which is the standard algebraic symbol for "change in".(8 votes)
What do you do when you don't have the y-intercept?(2 votes)
What about when you don't have the y-intercept? Not for the linear function, but the exponential one?(2 votes)
- I dont get this can someone please help(1 vote)
Hey, so I came up with a good way to remember how to do this type of problem, so I'll give you an example to demonstrate.
x | f(x)
---------
0 | 9
1 | 15
So we have the first number, 9, become the second number, 15. a good trick to remember is if it is on the place where x=1, put that on the top. I call this "The Big Switcheroo." If it is on the top, put it on the bottom. So if we switch the 9 and the 15, we will get y=15/9^x. Any Questions?(3 votes)
how do we write a exponential equations(2 votes)
An exponential equation is an equation that has a x as the exponent. If when x is 0 and y is not 1, that means there is a multiplier. The multiplier is equal to the y-intercept.
If the y-value is decreasing, then it's called exponential decay and the base of x must be smaller than 1.(1 vote)
I don't get something: Why did Sal do 'Δf/Δx' instead of just solving 'mx + b' for 'x' like 'm1 + 5 = 7', '(2)1 + 5 = 7', so 'm = 2'?(1 vote)- The use of
Δf/Δxis the standard form of finding the slope. It is good for those times when you don't know the value of x and y.(3 votes)
1:30Video transcript
Consider the following table
of values for a linear function f of x is equal to mx plus b
and an exponential function g of x is equal to a
times r to the x. Write the equation
for each function. And so they give us,
for each x-value, what f of x is and
what g of x is. And we need to figure out the
equation for each function and type them in over here. So I copy and
pasted this problem on my little scratchpad. So let's first think
about the linear function. And to figure out the
equation of a line or a linear function
right over here, you really just need two points. And I always like to use the
situation when x equals 0 because that makes
it very clear what the y-intercept is going to be. So, for example, we
can say that f of 0 is going to be equal
to m times 0 plus b. Or this is just going
to be equal to b. And they tell us that
f of 0 is equal to 5. b is equal to 5. So we immediately know
that this b right over here is equal to 5. Now, we just have
to figure out the m. We have to figure out
the slope of this line. So just as a little bit
of a refresher on slope, the slope of this line is
going to be our change in y-- or our change in our function
I guess we could say, if we say that this
y is equal to f of x-- over our change in x. And actually, let me
write it that way. We could write
this as our change in our function
over our change in x if you want to look
at it that way. So let's look at
this first change in x when x goes from 0 to 1. So we finish at 1. We started at 0. And f of x finishes
at 7 and started at 5. So when x is 1, f of x is 7. When x is 0, f of x is 5. And we get a change
in our function of 2 when x changes by 1. So our m is equal to 2. And you see that. When x increases by 1, our
function increases by 2. So now we know the
equation for f of x. f of x is going to be equal
to 2 times 2x plus b, or 5. So we figured out
what f of x is. Now we need to figure
out what g of x is. So g of x is an
exponential function. And there's really two things
that we need to figure out. We need to figure
out what a is, and we need to figure out what r is. And let me just rewrite that. So we know that g of x--
maybe I'll do it down here. g of x is equal to a
times r to the x power. And if we know what g of 0 is,
that's a pretty useful thing. Because r to the 0th
power, regardless of what r is-- or
I guess we could assume that r is not equal to 0. People can debate what
0 to the 0 power is. But if r is any
non-zero number, we know that if you raise it
to the 0 power, you get 1. And so that
essentially gives us a. So let's just write that down. g of 0 is a times r to
the 0 power, which is just going to be equal
to a times 1 or a. And they tell us what g of
0 is. g of 0 is equal to 3. So we know that a is equal to 3. So so far, we know that our g
of x can be written as 3 times r to the x power. So now we can just use any
one of the other values they gave us to solve for r. For example, they tell us
that g of 1 is equal to 2. So let's write
that down. g of 1, which would be 3 times r to
the first power, or just 3-- let me just write it. It could be 3 times
r to the first power, or we could just write
that as 3 times r. They tell us that g
of 1 is equal to 2. So we get 3 times
r is equal to 2. Or we get that r
is equal to 2/3. Divide both sides of
this equation by 3. So r is 2/3. And we're done. g of x is equal to 3 times 2/3. Actually, let me just
write it this way. 3 times 2/3 to the x power. You could write it that way
if you want, any which way. So 3 times 2/3 to the x power,
and f of x is 2x plus 5. So let's actually
just type that in. So f of x is 2x plus 5. And we can verify that that's
the expression that we want. And g of x is 3 times 2
over 3 to the x power. And let me just verify that
that's what I did there. I have a short memory. All right. Yeah, that looks right. All right. Let's check our answer. And we got it right.