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## Algebra 2

### Course: Algebra 2 > Unit 9

Lesson 7: Graphs of exponential functions# Graphing exponential functions

CCSS.Math: ,

Sal graphs y=-2*3ˣ+5 using our interactive graphing widget.

## Want to join the conversation?

- I csnt really understand when should the asymptote be parallel to x axis and when will it be parallel to y axis. This is one of the hardest math lessons i had ever seen(10 votes)
- An asymptote that is parallel to the x-axis is a horizontal asymptote. An asymptote that is parallel to the y-axis is a vertical asymptote.(3 votes)

- I don't understand why will y approach 5 as x gets negative?(4 votes)
- 0:08-1:00

We won't know how a graph is shaped until we know how it responds to certain values of x. (This is why it's good to use a table when working on graphs.)

Here we have y = -2 * 3^x + 5.

We can use a table to find the graph's shape, but we can also observe what happens to the graph as x approaches positive or negative infinity.

3 to the power of super negative numbers almost equals zero, so -2 * 3(-oo) will be a number*slightly*less than zero. Numbers like this, plus the remaining 5, will approach 5.(4 votes)

- but doesn't the ( -2) indicate that the new function must be reflected across the x-axis? i see no reflections at all..(4 votes)
- Yes, the -2 indicates that the new function must be reflected across the x-axis, but what your forgetting is that this is a transformation from the parent graph, in this case, is the exponential graph. If you look at the standard exponential graph, it basically looks like this graph but flipped over the x-axis and a horizontal aymptote of 0 instead of 5. Hope this helps.(4 votes)

- Why wasn't it flipped over the x-axis and stretched vertically by a factor of -2? and what does it mean to stretch vertically by a factor of a number? for example in this case it would be stretched vertically over the x-axis by a factor of -2. There are problems in the practice that say that(5 votes)
- Can someone help me understand how to graph a function where y=f^z-x? I understand that I'm supposed to start by rewriting them as y=f^(-(x-z)), but every time I attempt to graph those specific functions, I always seem to get it wrong.(3 votes)
- Ok, so y = f^(-(x-z)) can be graphed quite simply with a few rules to follow. Firstly, the sign in front of the exponent (which is - in this case) means that it is reflected upon the y axis. As for -z, the function is shifted z units towards the right. Remember, you always want the function to be zero, which means if you have z, you move -z units left. If you have -z, you move z units right. I hope this helped!(3 votes)

- In {practice:Graphs of exponential } there's a point I don't understand : What's the different between 3^5-x and 3^x-5 ?For 3^x-5, in oder to make y equal to one , x need to be equal to 5, so I transform it 5 to the right.But in 3^5-x, in oder to make y equal to one, x also need to be equal to 5!So again I transform it 5 to the right!So isn't that means:

3^5-x = 3^x-5 ?(3 votes) - Would the range be y>(the asymptote) and also how would we figure out the anchor points to graph?(2 votes)
- Since you have a -2 as a multiplier, it reflects across x, so the range would be y< (asymptote). O and 1 as x values are generally good points unless there is a horizontal shift (due to channging x such as y = -2 (3)^(x-2) which moves equation 2 units ot the right, this would mean x values such as 1, 2, and/or 3 would be good points(2 votes)

- I don't get how the first part(the left side) is going to approach zero?(2 votes)
- The left part of the side is approaching y=5. That is your horizontal asymptote — when x is
*approaching*-infinity (as you can see in the graph), it will hit y=5.

In the form of a(b)ˣ+c, c is your horizontal asymptote. Naturally, heres a link regarding this topic:

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:exponential-growth-decay/x2f8bb11595b61c86:exponential-vs-linear-growth/v/exponential-growth-functions

For your second part in the comments, he got -1 by plugging in 1 in the x in the equation. Looking at the equation:

f(x) = -2(3)ˣ+5

f(1) = -2(3)¹+5

f(1) = -6+5

f(1) = -1

Horizontal (and later vertical) asymptotes can be very tricky, so I recommend practicing these types of questions.(2 votes)

- Can someone also please explain the transformations in this exponential function compared to its parent function of 2^x? Thanks!(2 votes)
- If we take the parent function to be y = 2^x, then the function in question can be expressed as y = -2*2^(x*log_2(3)) + 5. So, our function can be considered as the result of the following transformations on its parent function y = 2^x:

1. Multiplying the function input by log_2(3) causes a horizontal scaling by the factor of log_2(3). And since |log_2(3)| > 1, the horizontal scaling in this case is a horizontal compression.

2. Multiplying the function output by -2 causes two things. A vertical scaling by the factor of |-2| = 2 and vertical reflection across the asymptote since the scaling factor is negative.

3. Adding 5 to the function output causes vertical shift upwards by 5 units (the asymptote becomes y = 5).(2 votes)

- How do we determine the horizontal and vertical asymptote for all the exponential function?(2 votes)
- exponential functions do not have a vertical asymptote. let's look at a simple one first though.

2^x

So obviously the horizontal asymptote is 0. Now, there are four things we can do to transform it.

If you multiply outside of the function, like 3*2^x this does not effect the horizontal asyptote (which I will call HA for now). Worth noting if you make the number negative it flips it over the x axis, though the HA will still be 0.

Multiplying inside the function doesn't do anything either, so something like 2^(-3x), it flips the graph over the y axis since it's negative, but the HA stays that same.

Let's try adding. 2^(x+3) moves all points to the left by 3 while 2^(x-3) would move all points to the right by 3. moving the HA left or right doesn't do anything to it.

If you add like this though 2^x + 3 it moves everything up by 3, including the HA. So adding outside of the function will move the HA, while subtracting of course will move it down.

I there si anything you didn't understand let me know, or if you have any more questions.(2 votes)

## Video transcript

- [Voiceover] We're told, use
the interactive graph below to sketch a graph of y
is equal to negative two, times three to the x, plus five. And so this is clearly
an exponential function right over here. Let's think about the behavior as x is, when x is very negative, or when x is very positive. When x is very negative, three
to a very negative number like let's say you had three
to the negative 3rd power, that would be 1/27th, or three
to the negative 4th power, that would be 1/81st. So this is going to get smaller,
and smaller, and smaller. It's going to approach zero
as x becomes more negative. And since this is approaching zero, this whole thing right over
here is going to approach zero. And so this whole expression is, if this first part is approaching zero, then this whole expression
is going to approach five. So we're going to have
a horizontal asymptote that we're going to approach
as we go to the left. As x gets more and more negative, we're going to approach positive five. And then as x gets larger,
and larger, and larger, three to the x is growing exponentially. But then we're multiplying
it times negative two, so it's going to become more,
and more, and more negative. And then we add a five. And so what we have here,
well this looks like a line. We want to graph an exponential. So let's go pick the
exponential in terms of x. There you have it. And so we can move three things. We can move this point,
it doesn't even need to just be the y-intercept, although that's a convenient
thing to figure it out. We can move this point here. And we can move the asymptote. And maybe the asymptote's
the first interesting thing. We said as x becomes
more, and more, and more, and more negative, y is
going to approach five. So let me put this up here. So that's our asymptote. It doesn't look like it quite yet, but when we try out some values for x and the corresponding y's, and we move these points accordingly, hopefully our exponential
is going to look right. So let's think about, let's
pick some convenient x's. So let's think about
when x is equal to zero. If x is equal to zero, three
to the zeroth power is one. Negative two times one is negative two, plus three is three. So when x is equal to zero, y is three. And let's think about
when x is equal to one, and I'm just picking that
'cause it's easy to compute. Three to the first power is
three, times negative two is negative six, plus
five is negative one. So when x is one, y is negative one. And so let's see, does
this, is this consistent with what we just described? When x is very negative,
we should be approaching, we should be approaching positive five, and that looks like the case. As we move to the left, we're
getting closer, and closer, and closer to five, in fact,
it looks like they overlap, but it's really, we're
just getting closer, and closer, and closer 'cause this term, this term right over
here is getting smaller, and smaller, and smaller
as x becomes more, and more, and more negative. But then as x becomes
more and more positive, this term becomes really negative 'cause we're multiplying
it times a negative two, and we see that it
becomes really negative, so I feel pretty good about
what we've just graphed. We've graphed the horizontal
asymptote, it makes sense, and we've picked two
points that sit on this, on the graph of this exponential. So I can check my answer,
and we got it right.