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### Course: Integrated math 3 > Unit 6

Lesson 7: Graphs of exponential functions# Transforming exponential graphs

Given the graph of y=2ˣ, Sal graphs y=2⁻ˣ-5, which is a horizontal reflection and shift of y=2ˣ.

## Want to join the conversation?

- I'm a bit confused here..because 2^2=4 yes BUT 2^-2=1/4 Or perhaps I'm just not understanding this problem fully.. Please help out.(54 votes)
- What you just said, is what Sal graphed!

With a x value of -2, the y value becomes 4.

2^-(-2) = 2^(2) = 4

hence the coordinate, (-2, 4)

With a x value of 2, the y value is 1/4, or 0.25.

2^-(2) = 2^-2 = 1/4

the coordinate is (2, 1/4).(58 votes)

- How did Sal know that -5 was going to be the horizontal asymptote, why not the vertical one?(7 votes)
- Because we know the graph of y=2^x has a horizontal asymptote as y=0

The graph y=2^(-x) reflects y=2^x over the y-axis

y=2^(-x)-5, the -5 is the vertical shift, so it moves the graph 5 units down. Essentially, it moves the horizontal asymptote 5 units down as well.(13 votes)

- Anyone have any idea about what "asymptote" means at1:53?(10 votes)
- An asymptote is an 'imaginary' line, that the curve/function approaches but never touches (nor intersect).(10 votes)

- I think the explanation isn't very helpful here... At0:24, Sal says "Any input we now put into x, we now take the negative of it, so if I input a 2, it's like taking the opposite of the 2 and then inputting that into 2^x." - and then he immediately goes "So it's like we're flipping the graph over the y-axis."

I feel like that's quite a mental jump that isn't at all immediately intuitive to me (and it appears others as well, judging from the comments).

2^(-2) = 1/4 and if we set y to 4, we get this at 4 = 2^-(-2) = 2^(2).

So while I can see numerically that we're flipping the graph horizontally, Sal's explanation isn't intuitive to me at all here, and seems to be making too big a leap.(7 votes) - why is not the value of y = 1/4 ,since the exponent is = -2(3 votes)
- Could you please specify a time stamp or reference point in the video? Sal is working with multiple different functions and graphs. All involve 2^x or 2^(-x). So, it is unclear what you are referring to.(3 votes)

- Is mathematics 3 a high school level or university ?(1 vote)
- Mathematics 3 is high school level mathematics.

In most universities they would expect that you understand most of the topics in that section.(4 votes)

- How can i do this algebraically?(2 votes)
- How did one get the equation for exponential functions from f(x) = a(k(x-d)) + c to f(x)= a ^k(x-d) + c?(1 vote)
- Is Mathematics III apart of Algebra?(1 vote)
- Yes. In fact, Math III contains mainly Algebra II topics. There is not a lot of geometry.(1 vote)

- The equation a(b)^2+c is the original exponential equation. can someone list all the possible changes there are.

ex: a is negative so the graph will transform to.......

ex: b is less than 1 so.......

ex: c is 1 so..............(1 vote)

## Video transcript

- [Voiceover] We're told the graph of y equals two to the x is
shown below, alright. Which of the following is the graph of y equals two to the
negative x minus five? So there's two changes here. Instead of two to the x, we
have two to the negative x and then, we're not leaving that alone, we, then, subtract five. So let's take 'em step by step. So let's first think about what y equals two to the negative
x would look like. Well, any input we now put into an x, we're now going to take the negative of. So if I input a two, it's
like taking the opposite of the two and then, inputting
that into two to the x. And so, what we're essentially going to do is flip this graph over the y-axis. So, here we have the point two comma four. Over here, we're going to have the point negative two comma four. When x is zero, they're going
to give us the same value. So they're both going to
have the same y-intercept. And so are graph is going to look like, our graph is going to
look something like, this. They're going to be mirror images flipped around the y-axis. So, it's going to look like that. That is the graph of y is
equal to two to the negative x. And then we have to worry about the subtracting five from it. Well, that's, you're subtracting five from your final y-value
so that's going to work. You're subtracting five
to get your y-value now, where your y-value is
going to be five lower, is I guess the best way to say it, so this is going to shift
the graph down by five. So instead of having
the y-intercept there, it's going to be five lower. One, two, one, each hash mark is two so this is one, two, three, four, five, is going to be right over there. So shift down by five, two, four, five. It's going to look like that. And the asymptote,
instead of the asymptote, going towards y equals zero, the asymptote is going to be at y is
equal to negative five. So the asymptote is going to
be y equals negative five. So it should look something like, something like what I
am drawing right now. Something like, something like that. So now we can look at which choices. So this should be the graph of y equals two to the negative x minus five. So let's see which of
these choices depict that. So this first choice
actually seems to be spot-on. It's exactly what we drew. Well, we can look at the
other ones, just in case. Well, this is, this looks like,
what did they do over here? Looks like they, instead of flipping over the y-axis, they took the,
they flipped over the x-axis and then they shifted
down, so that's not right. Here it looks like they got what we got but they flipped it over the x-axis. And this looks like they
flipped it over the y-axis but then they shifted,
instead of shifting down by five, it looks like they
shifted to the left by five. So we should feel pretty good especially 'cause we essentially drew this before even looking at the choice.