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## Integrated math 3

### Course: Integrated math 3>Unit 6

Lesson 8: Graphs of logarithmic functions

# Graphing logarithmic functions (example 2)

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? • 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? • 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!
• 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? • At , 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? • 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. • 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.
• Hi!
Why is the asymptote of the function on the graph at -6 , shouldn't it be on -7 as given in the equation ? • From what I've gathered from the previous videos and practice exercises:
>>>"2^x functions" have horizontal asymptotes
>>>"log_2(x) functions" have vertical asymptotes

Is this always true or is there more learn? • 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) • 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) 