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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

# Graphing exponential functions

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 • I don't understand why will y approach 5 as x gets negative? • -

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.
• but doesn't the ( -2) indicate that the new function must be reflected across the x-axis? i see no reflections at all.. • 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.
• 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 • 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. • 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!
• 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 ? • Would the range be y>(the asymptote) and also how would we figure out the anchor points to graph? • I don't get how the first part(the left side) is going to approach zero? • Can someone also please explain the transformations in this exponential function compared to its parent function of 2^x? Thanks! • 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).
• How do we determine the horizontal and vertical asymptote for all the exponential function? • 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.