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## Algebra 2

### Course: Algebra 2 > Unit 9

Lesson 8: Graphs of logarithmic functions# Graphing logarithmic functions (example 2)

CCSS.Math: ,

We can graph y=4log₂(x+6)-7 by viewing it as a transformation of y=log₂(x).

## Want to join the conversation?

- How do I graph the function from scratch without a graph initially? As in just given a blank graph and f(x)= 2 log_4 (x+3)-2? Also, what is the 2 in the front and the -2 at the end of the function?(9 votes)
- The 2 in front means that the log means that the logs y value is multiplied by 2. The -2 at the end of the log means the graph is shifted 2 down.

To graph the function from scratch without a graph would mean that you would have to memorise the graph.

Hope this helps.(4 votes)

- I've been trying this for a while, but I don't understand how you evaluate where the asymptote would be from an equation. When is the disconnected number (separate from the x expression) the asymptote and when do you set the entire equation to zero?(6 votes)
- Since log(0) is basically asking what you would raise some number to to get 0, it is undefined, as no exponent by itself can get you to 0. So, to find the vertical asymptote, we must look for the point at which the part inside the logarithm (its argument) would be 0. Since the asmptote is vertical, you only need to look at the horizontal transformations to determine its location. Set the argument to 0, and solve.

As an example, let's take f(x) = log_2(3(x^2 - 9)) + 9

From this, we get 3(x^2 - 9) = 0

= (x^2 - 9)

= (x + 3) (x - 3)

x = 3, -3

So, the vertical asymptotes for this function are 3 and -3. Hope this helps!(8 votes)

- Can someone please explain what you do when you have something like -log2? How would you transform the graph for the negative that is in front of the log?(4 votes)
- When you put the negative in front of the function, that means that you are reflecting it across the x-axis.(5 votes)

- At0:13, Sal says log base 2 (x) has a vertical asymptote at zero, and has the points 1,0 and 2,1 highlighted, but in the exercise it has the points 1,0 and 5,5 highlighted at the start. Is it just a glitch, or what?(4 votes)
- The dashed line has (1,0) and (2,1) highlighted. The point of the exercise was to make the part you could move match it. The part that moves has the point (4,4) highlighted.(1 vote)

- I know Adrianna already asked this question but I don't find Timothy_Lavrenov's answer satisfying. How do I graph a log function without having seen anything else? Like a piece of graphing paper and a log function. Now graph.(2 votes)
- log functions do not have many easy points to graph, so log functions are easier to sketch (rough graph) tban to actually graph them. You first need to understand what the parent log function looks like which is y=log (x). It has a vertical asymptote at x=0, goes through points (1,0) and (10,1). With a lot of graphs, you will not even be able to reach the (10,1) point if you are moving it around. Next, you need to know your transformations which are relative to all functions f(x) = a f(bx+c)+d. A is a vertical stretch or compression as well as reflect across x if negative, b is a horizontal stretch or compression as well as having a negative reflect across y. C are translations to the left and right, and d shifts up and down. With the parent function, you would draw the horizontal asymptote at x=0, plot the points (1,0) and (10,1) and draw a rough curve. Given the function of Adrianna f(x)=2 log(x+3)-2, the transformations to the parent function would include a vertical stretch and a shift of (0,0) to (-3,-2) which you then act as if it is (0,0) even though it really is not. This gives a vertical asymptote at x=-3 which is the start. With a shift down 2 and a multiplier of 2 (vertical stretch). Then at (-3,-2), you would still move <1,0> to get to (-2,-2), but instead of <10,1>, it would move <10,2> (notice in both cases, you just multiplied the y value by 2) to get to (-3+10, -2+2) or (7,0). Hope this is a little more satisfying to you.(3 votes)

- Hi!

Why is the asymptote of the function on the graph at -6 , shouldn't it be on -7 as given in the equation ?(2 votes)- -7 is the horizontal asymptote and +6 is the positive asymptote - the one outside the parenthesis is the horizontal asymptote. Hope this helped!(2 votes)

- I understand how to do the problems in the first two videos, and do overall understand graphing logs, but I had a question about what to do if that 4 were negative. Is it a reflection and a stretch by four? or would that turn it into a compression by 4 (1/4)(1 vote)
- If you made the 4 negative, it would be a reflection across y = -7. The reason, as you might be able to tell, is that pesky -7 at the end of the function. That doesn't get negated, so everything revolves around it.(1 vote)

- How do I graph a function if the function is y=3log_2(-x)-9?(1 vote)
- Hi everyone,

In the video Sal operates y=2*log2(-(x+3); however, if we applied the rules of transformation to 2*log2(-x-3) we would have very different coordinates of the graph.

Or let's take another one; y=-3*log3(1-x)+6

I do not know whether I should solve it as y=-3*log3(-(x-1)+6 or

y=-3*log3(1-x)+6. It results into two different coordinates of the graph.

I experience similar difficulties with transformations of the radical graphs. Can I treat y=3√(2−x ) or it is better to ask should i handle it as

3√(−(x−2) )?

I'm asking this because the equation y=3√(2−x) is transformed by reflecting it across y-axis , shifting the function two to the left and multiplying y coordinate values by 3, but it produces the answer that doesn't match the solution on Khan Academy. Should I shift it two to the right and multiply

Y-coordinate values by 3 ---> 3√(−(x−2) )? How can i be sure whether transform it as 3√(−(x−2) ) or as y=3√(2−x ) ?

The solution of the third question is (-2,6) (0,4) (2,0)

Thank you.(1 vote)

## Video transcript

- [Instructor] This is a screenshot from an exercise on Khan Academy, and it says the intergraphic,
the interactive graph below contains the graph of y is equal to log base two of x as a dashed curve, and you can see it down
there as a dashed curve, with the points one comma zero and two comma one highlighted. Adjust the movable
graph to draw y is equal to four times log base two
of x plus six minus seven. And so if you happen to have
this exercise in front of you I encourage you to do that. Or if you're just thinking
about in your head, think about how you would approach this. And I'll give you a hint,
to go from our original y is equal to log base
two of x to all of this, it's really going to be a
series of transformations. And on this tool right over here, what we can do is we can move
this vertical asymptote around so that's one thing we can move, and then we can also
move two of these points. So where we're starting is right, we are starting right over there. And so let's see, and
that was just the graph of y is equal to log base two of x. So let's just do these
transformations one at a time. So the first thing I am going to do, instead of just doing log base two of x, let's do log base two of x plus six. So if you replace your
x with an x plus six, what is it going to do? Well it's going to shift
everything six to the left, and if that doesn't make
intuitive sense to you, I encourage you to watch some of the introductory videos
on shifting transformations. So everything is going
to shift six to the left. So this vertical asymptote is
going to shift six to the left it's gonna be, instead of
being at x equals zero, it's going to go all the way
to x equals negative six. This point right over here,
which was at one comma zero, it's going to go six to the left, one, two, three, four, five, six. And this point, which as at two comma one, is gonna go six to the left, one, two, three, four, five, and six. So so far what we have graphed is log base two of x plus six. So the next thing we might wanna do is what is four time log
base two of x plus six. And I want you to think about it is whatever y-value we were getting before, we're now going to get four times that. So when x is equal to negative five, we're getting a y-value of zero, but four times zero is still zero, so that point will stay the same. But when x is equal to negative four, we're getting a y-value of one, but now that's going to
be four times higher, 'cause we're putting that four out front, so instead of being at four, instead of being at one
it's going to be at four. So this right over here is the graph of y is equal to log
base two of x plus six. And then the last thing
we have to consider is well we're gonna take all of that and then we're going to subtract seven to get to our target graph. So whatever points we are here, we are now going to subtract seven. So this is at y equals zero, but now we're going to subtract
seven, so we're going to go down one, two, three,
four, five, six, seven, I went off the screen a
little bit, but let me see if I can scroll down a little
bit so that you can see that, almost, there you go, now you can see. I moved this down from
zero to negative seven, and then this one I
have to move down seven, one, two, three, four,
five, six, and seven, and we're done, there you have it. That is the graph of y is equal
to four times log base two of x plus six minus
seven, and we are done.