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## Algebra 2 (Eureka Math/EngageNY)

### Course: Algebra 2 (Eureka Math/EngageNY) > Unit 3

Lesson 9: Topic C: Lessons 17-22: Graphs of exponential and logarithmic functions- Relationship between exponentials & logarithms
- Relationship between exponentials & logarithms: graphs
- Relationship between exponentials & logarithms: tables
- Relationship between exponentials & logarithms
- Transforming exponential graphs
- Transforming exponential graphs (example 2)
- Graphing exponential functions
- Graphs of exponential functions
- Graphical relationship between 2ˣ and log₂(x)
- Shape of a logarithmic parent graph
- Graphs of logarithmic functions
- Graphs of logarithmic functions

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# Transforming exponential graphs (example 2)

CCSS.Math: ,

Given the graph of y=2ˣ, Sal graphs y=(-1)2ˣ⁺³+4, which is a vertical reflection and a shift of y=2ˣ.

## Want to join the conversation?

- what is the difference between Geometric sequence equation and Exponential equation.(8 votes)
- Although both multiply by a common ratio to get to the next number, there are still some subtle differences between them. Geometric sequences are exactly that:
*sequences*. They are a never-ending set of numbers*in a list*. An exponential equation can only be represented by a graph, and you can't put those values in a list (there are just too many).(9 votes)

- Shall I first learn graphs before getting into this story?(3 votes)
- Yes, start with how to graph points, then graphing linear functions, then graphing quadratic functions. Once you understand those, you should be okay for exponential graphs.(15 votes)

- when Sal says y=2 to the power of x+3, i got confused. How did he get the power of x+3 simplified? What steps did he use?(4 votes)
- Sal had the graph y = 2^x.

The second graph, y = -1 * 2^(x+3) + 4, is a transformation of y = 2^x.

According to March11, Sal must use PEMDAS to track how y = 2^x will transform.

He first works with the exponent (E).

A 3 was added to the x in y = 2^x, making it y = 2^(x+3). Sal transforms the graph accordingly. Afterwards he moves on to the other parts of the transformation (multiplying -1 and adding 4).

There is no simplification involved, only working by chunks.(3 votes)

- In the first graph that Sal made,
`y=2^x+3`

, at2:20. Why is the y intercept 5? I thought that if x is 0 then 2 would be to the power of 3 which is 8, right? or is it a error?(2 votes)- As drawn by Sal, the y-intercept is 6. However, you are correct that it should be 8. The difference comes because Sal's drawing is only an approximate sketch. Further on in the video at about4:11, there is a pop-up note that also mentions another inaccuracy of Sal's sketch.(7 votes)

- Is the constant always going to be the horizontal asymptote?(4 votes)
- No, not always.

Consider the rational function*f(x) = (ax^n + bx + 6) / (cx^m + dx + 6).*

The horizontal asymptote is the horizontal line that*f(x)*approaches as x approaches positive/negative infinity. You don't have enough information to find that, but you do have enough information to find the y-intercept. What is*f(x)*when x is zero?

f(0) = 6/6 = 1

y-intercept: (0, 1)

Constants will not be an HA if you're also working with linear equations.

I'm not sure how else constants will be the HA, but it would be a good topic to research.(3 votes)

- i cant understand why in y=2^-x we flipped the graph over the y-axis and in y=-1x2^x+3 we flipped it over x-axis(3 votes)
- Transformation on the basis of reflection follow these two rules:

f(-x) = Reflection off y axis

-f(x) = Reflection off x axis

I recommend watching this video about it if you want a visual representation, or head to desmos:

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:transformations/x2ec2f6f830c9fb89:reflect/v/reflecting-functions-examples

Hopefully that helps !(4 votes)

- How do you determine the y-intercept of an exponential function f(x)= ab^x + k that has been both stretched and translated?(3 votes)
- For any equation, you find the y-intercept by using x=0.

Hope this helps.(4 votes)

- Does anyone know of a faster way to find x values that work to plot? I keep going through all of the x values and solving for them, which is extremely time consuming and annoying if I solve for the correct plot-able x value incorrectly and have to start all over again.(3 votes)
- If we are multiplying the original equation by -1 at0:38, why isn't there a parenthesis on it. Doesn't that affect something?(4 votes)
- Can someone help me describe the transformations of y= -2(3^-x)+1(2 votes)
- The parent graph is y = 3^x

The transformations in order: reflection in y-axis, followed by reflection in x-axis, followed by vertical stretch by a factor 2, and finally a vertical shift upwards of 1 unit.(4 votes)

## Video transcript

- [Voiceover] We're told the graph of y equals two to the x is shown below, so that's the graph. It's an exponential function. Which of the following is a graph of y is equal to negative one times two to the x plus three plus four? They give us four choices down here. Before we even look
closely at those choices, let's just think about what this would look like if it was
transformed into that. You might notice that what we have here, this y that we wanna find the graph of, is a transformation of this original one. How do we transform it? We've replaced x with x plus three. Then we multiplied that by negative one, and then we add four. Let's take it step by step. This is y equals two to the x. What I wanna do next is let's graph y is equal to two to the x plus three power. If you replace x with x plus three, you're going to shift the graph to the left by three. That might be a little
bit counterintuitive, but when we actually
think about some points, it'll hopefully make some sense here. For example, over in our original graph, when x is equal to zero, y is equal to one. How do we get y equal
one for our new graph, for this thing right over here? To get y equals one
here, the exponent here still has to be zero, so
that's going to happen at x equals negative three. That's going to happen at
x equals negative three. Y is equal to one. Notice, we shifted to the left by three. Likewise, in our original graph, when x is two, y is four. How do we get y equals four in this thing right over here? For y to be equal to
four, this exponent here needs to be equal to two,
and so for this exponent to be equal to two, 'cause
two squared is four, for this exponent to be equal to two, x is going to be equal to negative one. When x is equal to negative one, y is equal to four. When x is equal to negative one, y is equal to four. Notice we shifted to the left by three. So this thing, which isn't our final graph that we're looking for, is gonna look something like, like that, which shifted,
it's y equals two to the x, shifted to the left by three. Now let's figure out what the graph of, now let's multiply this
expression times negative one. Notice we're slowly building up to our goal. Now let's figure out the graph of y is equal to negative one times two to the x plus three. Here, when y equals two
to the x plus three, if we multiply that times negative one, whatever y we had, we're gonna have the negative of that. Instead of when x is
equal to negative three having positive one, when
x equals negative three, you're gonna have negative one. We multiplied by negative one. When x is equal to negative
one, instead of having four, you're going to have negative four. Our graph is gonna be flipped over, it's flipped over the x axis. It's going to look something like this. This is not a perfect
drawing, but it'll give us a sense of things, and we can look at which of these graphs match up to that. Then finally, we wanna
add that four there. We wanna figure out the graph of y equals negative one times two to the x plus three plus four. We wanna take what we just had and shift it up by four. Instead of this being a
negative one right over here, this is going to be a negative one plus four is three. Instead of this being a negative four, negative four plus four is zero. Instead of our horizontal asymptote being at y equals zero,
our horizontal asymptote is going be at y equals four. It's gonna look like, let me draw, I can do a better job than that. Our horizontal asymptote is gonna be right over there, so our graph is going to look something like this. We just shifted that red graph up by four. Shifted it up by four. We have a horizontal
asymptote at y equals four. Let's look at which of
these choices match that. Choice A right over here
has a horizontal asymptote at y equals four, but it is shifted on the horizontal
direction inappropriately. In fact, it looks like it might have not been shifted to the left. We can rule this one out. Let's rule that one out. This one over here, this one approaches our
asymptote as x increases, so that's not right. It should approach our asymptote as x decreases, so we ruled that one out. Choice C looks like what we just graphed. Horizontal asymptote at x equals four. When x equals negative
three, y is equal to three. That's what we got. When x is equal to negative one, y is equal to zero. This looks right. You could even try those points out. We like choice C. D is clearly off.