HELLO, Mr. Tarrou Today we are doing a video introducing transformations. Specifically, we are going to pick apart this parent function of y equals the square root of x. We we are going to take this parent function y equals the square root of x and see what this function likes like as it sits right now. Lets make a t-table with this value (equation) and use some values for x like 0, 4, 9, 16. Lets see what this looks like graphically. I am using perfect squares because when I take those values and plug them in for x I am going to square root them. Why not pick numbers that I can graph easily. I don’t think I am going to use my x value of 16 in this small grid. We will go along with it anyway. If I plug in zero for x, I get the square root of zero is zero. Plugging in 4, I have the square root of four which is two. If I plug in nine, the square root of nine is equal to three. And not that I have room to graph an x value of 16, the square root of sixteen is four. So this is going to look like (0,0) , over four and up two, over nine and up three. So 1,2,3,4,5,6,7,8,9 and up three. Here we have our parent function y equals the square root of x I don’t want to do a t-table every time I draw a graph, at least not when I am working with graphs I am already familiar with. If I want to graph y equals the square root of x minus 2, This plus or minus. When you have a plus or minus at the end of your equation (usually) that is outside of your function, in this case my function is the square root symbol, that is a vertical shift. If were plus two you would move the graph up, we have minus 2 so we will move the graph down. Each of these individual points I plotted a minute ago are going down two places because of this minus two. Now I can redraw my square root function shifted down two places. Again, plus or minus at the end of your equation (vertical shifts can be at the beginning) outside of your function, that is a vertical shift. Now what if I wanted to make this graph, y equals the square root of x minus two. A minus outside your function is a vertical shift. Vertical shifts are either up or down. If you have a plus or minus inside your function, in this case my square root symbol, that is a movement left or right. With a x minus two inside our function, that is a shift (translation) to the right two places. Not to the left two places like you might think because it subtraction, but it is to the right 2 units. If I were to plug in say… Lets set up a t-table. If I attempted to use a list of generic values such as zero, like this graph that starts at zero, zero minus two is negative two. You cannot square root negative two. You cannot square root a negative number. You can only square root a positive number. If I put in a value of two, two minus two is zero and the square root of zero is zero. You see how this point is at (2,0) where it used to be at (0,0) We are going to go over here and plot (2,0) Now lets put in 6. Why would I want to put in 6 when you cannot square root six? The first thing you are going to do is subtract by two, so six minus two is equal to four. Oh, the square root of four is equal to two. Originally my point was at (4,2), now to get that same point I have to move over to six. Six minus two is four and the square root of four is two. See these point are moving to the right two places to the right. You can see, if you have a minus to in your function that is going to be a shift (translation) to the right. If our equation was y equals the square root of x plus two, that would be a shift to the left. Lets take a look at another equation. I want to graph y equals negative square root of x. If you graph starts with a negative in front of the function, that indicates a reflection over the x axis. Lets look at this original white graph. My original graph, my parent function, y equals the square root of x at zero is zero. If I plug in zero, the square root of zero is zero and that negative sign does not change anything. My parent function at four gave me an answer of two. If I plug in four, the square root of four is two and now that negative out front is going to change my y value from two to negative two. Are you starting to see the pattern here? The graph is reflecting over the line y=0 or the x axis. If I go to the other x value I used of 9, the square root of nine was three. If I plug in nine of course the square root of that is still three, but there is that negative out front. What was once a positive three is now a negative three. If I have an equation, and I know my parent function, and it has a negative sign in front of the function that is going to be a reflection over the x axis. What was once above the x axis will be below the x axis. ALL RIGHTY! What if we take…..let me put that back up. How about we take the last one here before we move on to another example of y equals the square root of negative x minus two. I am going to look at my blue graph and compare these two. You may see this as Y=sqrt(-(x+2)) This form still shows the reflection about the y axis factored out, and more clearly shows the shift to the left in the parenthesis. How does y equals the square root of negative x minus two compare to the equation of y equals the square root of x minus two? The negative in front of the x is going to reflect the graph over the y axis. What was x minus two, what was starting at the x value of two and going to the right, what did have a horizontal shift to the right… now with this negative in front of the x INSIDE the function the graph is going to be reflected over the y axis. Instead of this point being at (2,0) like it was in this function, our new point is at (-2,0) With my blue graph because I am comparing the red with the blue, see they are almost the same except for the negative in front of the x INSIDE the function, what was at (6,2) will now be at (-6,2) The point at (11,3) will now be at (-11,3) So if you have again your parent function, a plus or minus at the end of your equation will be a vertical shift, a plus or minus inside your function will be a horizontal shift, if you have a negative in front of your function that is a reflection over the x axis, and if you have a negative in front of the x inside the function that is going to be a reflection over the y axis. So there we have a nice little graph that looks like spaghetti strings. Lets finalize this in one final statement again. Y is equal to plus or minus a. If you have a negative that is a reflection over the x axis. If a is greater than one, that is a vertical stretch. If a is greater than zero but less than 1 that is going to be a vertical compression. I did not do an example with that. I will do a stretch here in a second. Then we have this idea of f. If you textbook is using this notation, f is not a variable it represents a function. The function I just worked with in all those examples was the square root. This may be a set of parenthesis with a power of two like a quadratic, or even absolute value symbols. Inside that function you may have a plus or minus in front of the x. If you do have a negative here, that is going to be a reflection about the y axis. You may have a plus or minus C. If you have that plus or minus inside your function, then that is going to be a horizontal shift. If you see x plus c, that is a shift to the left c units. And if you see x minus c, that is a shift to the right c units. Plus or minus inside your function, that is a movement (translation) left or right. Finally you may have a plus or minus d and that will be a vertical shift. That is going to be our summary statement. You might also have a coefficient of B in here. My textbook does not deal with those much until we get to Trig functions. If you have a coefficient of B in here that is going to be a horizontal stretch or compression. I am not going to work with that until we graph trig functions and that is a couple of chapters away. ALRIGHT, one last example. When you are graphing something like say a parabola, the parent function for parabolas is y equals x squared might look something like this. Where as if you have something like y equals 2x squared, that two is a vertical stretch. If a is great that 1 it is going to pull the graph up. With parabola because they go up indefinitely it will look like they are skinnier. That can lead you into a little bit of trouble though with some of your more interesting graphs that you may have to deal with, say in a PreCalculus book. A vertical stretch does not really make a graph skinnier. Let me show you what I am talking about. If I give you a graph that is just a few line segments like this. We are going to call this function F. Now I want function G to equal two times function F. I want to put a vertical stretch on function F to create function G. That is going to take all the y values and multiply them by two. Lets take a look at this. This point has a y value of zero and two times zero is still zero. If I stretch this y value I end up here twice as far from the x axis. Now I have this increased slope on the side of my graph, like we see with the parabola. By fixing these corners down hopefully show that my function is not getting skinnier. All of these y values here are going to get twice as tall. You get a graph that is twice as far from the x axis. We are not getting that optical allusion that the graph is getting skinnier like we see in the parabola. Finally we come down to this y value of zero. Again zero times zero is zero and we are at the end of our graph. I have increased the slopes (steepness) of my sides, I have made my graph twice as tall with the stretch, but you can see by tying these ends down we did not make the graph skinnier. I just made it taller. That you do not see in a parabola because it goes up forever. That is transformations. I hope the video helped. BAM!!!

11:51 That "WOAH!" you just did was like you were talking for 5 minutes straight X'D

BAYUM

lol

Amazing. Nice work,

ty your a good, you helped me a ton with 12:30 –

oh my glob that you ur awesome i really loved it that i really got it!!! ðŸ˜€

BAMMM! Thanks for the help =)

Wow, the dotted lines, forming voxels, on which graph was initially drawn were also erased!!!! (at 9:27) They were drawn with chalk? How? They looked so perfect?

I love people who teach school online for free.

THANKS VERY MUCH SIR

Legend

These videos are awesome man! No joke. Its so nice that you breakdown how and why these rules and equation work when they're applied. I have read and reread my book several times and after watching this video it all makes sense now. haha.

Thanks for the help.

Hell yeah! thanks brother! BAM!!!

This the best video I have seen, you explained it fore the normal person to understand. Thank you!!!

GREAT video. But were you trying to see if we were still awake at 13:55?Â eeeek!Â Hahaha You have no worries of anyone falling asleep on your clearly explained lectures. Many of us owe our passing grades to you ðŸ˜‰

Amazing..truth beauty and goodness times infinity..or maybe to the nth power? ðŸ˜€

Music is far too loud

You taught this better than my teacher could teach it in a week

freaking awesome!

Hey ProfRob, This lecture of yours really helped a lot!! I was finally able to understand it, but if i get confused in my exam regarding the shifts of graph , is it okay if i make a small table including values of x and y of orginal function and then apply same in for the transformation, And Can you pls tell what to do if they give me coordinates of the curve and tell me to draw the transformation regarding the curve. Thanks ðŸ™‚

Better explanation than my lecturer ! thanks

You are one of the best teachers online! Pls keep up the great work!

now this is a quality vid but why not use a whiteboard maybe just me but i just cant stand the chalk scraping noise lol

VocÃª explica muito bem!

Is this Geometry or Trig?

You are the greatest to ever do this !!!!!!

Why wouldn't sqrt(x) a parabola that opens around the x axis

ALL RIGHT WOOO!!!! I love your enthusiasm ðŸ˜€

Right on! you're doing a great service for struggling students!!

Such an awesome professor, You saved me sir, thank you.

AWESOME!!! I have learned so much from your instructional videos over the past 6 months as a 35 yr old college student. They have helped me in maintaining a 4.0 GPA. Thank you so much! After watching some of your videos on topics I'm currently studying in precalc, I have come to a much fuller understanding of the concepts. I just made up a quadratic function, graphed it, and then produced a transformation for it and graphed it successfully. I've been watching your, and others', videos and studying, and practicing for the past six hours in preparation for my precalc exam tomorrow morning. I now feel more confident about it and am rather excited. Thanks again! You're awesome Mr. Tarrou!

yes preach robbob life saver

YOU ARE THE MAN!!!!!!!!!!!!!

i love you Mr. "Turoo" your videos are so helpful and the way you teach and speak make it easier to understand.

Thankyou sir , â˜º

Thank you so much, I was struggling with this section so much and your vid finally made a lot of things much clearer for me, I feel more confident for my quiz tomorrow

ill watch this anyday over 12 pages of notes

Taught me whole unit in 15 minutes.

Teacher takes 2 weeks and accomplishes nothingThanks

so helpfull omgg thank you

When I see an equation that I need/want to graph, my first reaction is to plug in positive values. But, in the case of graphing y=sqrt(-x-2), you end up with a complex number by doing that. So, I realized that if you plug in the neg values of the numbers used in y=sqrt(x-2), you get the reflection. Just an alternative that added to my understanding. I may not always recognize that it is a reflection. For instance, if I hadn't been given the first equation, and was told to graph y=sqrt(-x-2), I may not have even thought about reflection. I would have initially plugged in a positive number and been momentarily thrown off. Now I will always think of a way to make the equation under the radical a positive number. Thanks for the videos!

can you do this example :The graph of y = f(x) is shown. Sketch the graph of g (x) = â€“2f(2x + 2)

â€“ 3 on the same axis. State the domain and range of the transformed

function.

I seen your video mate, really love the content. Liked straight away, We should connect!

at around 9:20 isnt the points supposed to be shifted right and not left to -2?

All of your videos have helped me SO much!

thanks very much Prof, transformations tricked me a lot

You are awesome. Brian is right

Thank you for helping me understand this a little better than I have understood the past week I've done in class.

I really wish I could have you as my professor! You're amazing! I love your handwriting and your way of explaining things, makes everything seem so simple and easy when it has really not been!

thanks mr. seavey this guy is great

Finally I found someone to teach me in order to teach my son

Awesome videos for the IB Diploma! Thank you!

Your handwriting, sir, is so beautiful! Thank you for your help!!!:) gratitude from BC, Canada

I LOVE YOUR VIDEOS! I am back at math in college and I know these videos are going to help me out – ONCE AGAIN. THANK YOU! <3

Thank you sir. for some great explanations. I am very thankful that i found you on Youtube.

Im fascinated ! who draws your graphs lines?

So helpful!

Amazing lectures, thank you very much for your help!

Bam is right…..anything more on compressions and stretches? I"m confused about the parabola and the other box thing. If we make it twice as big in the second, but twice as small in the first, it confuses. What about a SIN function? ðŸ™‚

Thank you so much it seems like nobody else realized that when you stretch a parabola or absolute value graph vertically, it looks like you also compressed it horizontally

You explained it so well. Thank you!!

"BAM!"

This is great and fun!! Iâ€™m getting there! ðŸ¤—

One of the best sources

First of all, SO happy I found this video. Hopefully I will ace my quiz. Second, HE DREW THAT GRAPH BY HAND?! WHAT