Log in Destiny 10 years agoPosted 10 years ago. Direct link to Destiny's post “What is f(x) = |x| - 3T...” What is f(x) = |x| - 3 The fact that x is in between the absolute value sign confuses me. I know -3 would mean that we're going to the left on the horizontal plane, is that technically it? • (20 votes) Rachel 10 years agoPosted 10 years ago. Direct link to Rachel's post “f(x)=|x|-3. It's like f(x...” f(x)=|x|-3. It's like f(x)=x-3 except the 3 is inside absolute value brackets. The only difference is that you will take the absolute value of the number you plug into x. (37 votes) Ryujin Jakka 10 years agoPosted 10 years ago. Direct link to Ryujin Jakka's post “Are there more detailed v...” Are there more detailed videos that focus specifically on horizontal and vertical shifting and shrinking? Thanks • (11 votes) Lauren Edwardsen 6 years agoPosted 6 years ago. Direct link to Lauren Edwardsen's post “I use this reference form...” I use this reference formula g(x)=a*f((1/b)x-h)+k (15 votes) mdmoore37 4 years agoPosted 4 years ago. Direct link to mdmoore37's post “At 4:09, Why is it f(x-2)...” At • (9 votes) water613 2 years agoPosted 2 years ago. Direct link to water613's post “ayo did you figure it out...” ayo did you figure it out? cause i am wondered too (3 votes) Dontay Decker 4 years agoPosted 4 years ago. Direct link to Dontay Decker's post “What would the transforma...” What would the transformation do if g(x)=(x+6)^2-10 and g(x) is in absolute value bars? Like this: |g(x)|. • (6 votes) kubleeka 4 years agoPosted 4 years ago. Direct link to kubleeka's post “Taking the absolute value...” Taking the absolute value of a function reflects the negative parts over the x-axis, and leaves the positive parts unchanged. So a central segment of your parabola will be reflected so that it opens downward, with sharp corners at the roots. (4 votes) david haywood 4 years agoPosted 4 years ago. Direct link to david haywood's post “can some one help me? Wh...” can some one help me? • (3 votes) A/V 4 years agoPosted 4 years ago. Direct link to A/V's post “f(x)=x is equal to f(x)=x...” f(x)=x is equal to f(x)=x+0, just written in a more abstract way. This is useful when comparing to another linear functions such as your example. f(x)=x-5 is simply just f(x)=x brought down by 5 units, hence the "-5" for the b term. I recommend using desmos for a more visual interpretation, as writing down explanations can be more convoluted. Hopefully that helps ! (8 votes) Ramon M 7 years agoPosted 7 years ago. Direct link to Ramon M's post “Could anyone ennumerate a...” Could anyone ennumerate all the ways a function can be transformed? Thank you! • (4 votes) Jasmina Hasikic 7 years agoPosted 7 years ago. Direct link to Jasmina Hasikic's post “Well, a function can be t...” Well, a function can be transformed the same way any geometric figure can: (5 votes) Alexis313 5 years agoPosted 5 years ago. Direct link to Alexis313's post “f(x)=x,g(x)=x+1would the...” f(x)=x,g(x)=x+1 • (4 votes) loumast17 5 years agoPosted 5 years ago. Direct link to loumast17's post “Yep, for linear functions...” Yep, for linear functions of the form mx+b m will stretch or shrink the function (Or rotate depending on how you look at it) and b translates. Then if m is negative you can look at it as being flipped over the x axis OR the y axis. For all other functions, so powers, roots, logs, trig functions and everything else, here is what is hopefully an easy guide. a*f(b(x+c))+d so for example if f(x) is x^2 then the parts would be a(b(x+c))^2+d a will stretch the graph by a factor of a vertically. so 5*f(x) would make a point (2,3) into (2,15) and (5,7) would become (5,35) b will shrink the graph by a factor of 1/b horizontally, so for f(5x) a point (5,7) would become (1,3) and (10,11) would become (2,11) c translates left if positive and right if negative so f(x-3) would make (4,6) into (7,6) and (6,9) into (9,9) d translates up if positive and down if negative, so f(x)-8 would make the points (5,5) and (7,7) into (5,-3) and (7,-1) Also should note -a flips the graph around the x axis and -b flips the graph around the y axis. Hope I didn't over explain, just proud of what I made tbh (4 votes) Durgen a year agoPosted a year ago. Direct link to Durgen's post “I like how everyone is as...” I like how everyone is asking about certain math questions and the typical "where in real life would this be useful" kinda thing, and yet I seem to be the only one who's wondering about that fifth graph down at the bottom. What's that doing down there? Why did Sal not do any problems on that one but still did problems on the other four? These are legitimate questions. • (4 votes) Saubir21 a year agoPosted a year ago. Direct link to Saubir21's post “Maybe he had been plannin...” Maybe he had been planning to use it, but then he ran over time or something. Or it could be for another video. (2 votes) lonklomk 10 months agoPosted 10 months ago. Direct link to lonklomk's post “I have a really hard time...” I have a really hard time understanding if a graph is being compressed vertically or if it is being stretched horizontally and vice versa. You could make a case for both and the help feature on the exercise "Identify function transformations" does not explain why they came to that conclusion. • (4 votes) tran.48048 2 months agoPosted 2 months ago. Direct link to tran.48048's post “you can tell because a ho...” you can tell because a horizontal stretch wouldnt really change the y values on the graph but rather the locations that the y value is achieved for example, say there was a point (1,4) if this was stretched horizontally, the x value would change for example to (4,4) however you would notice that y=4 is still attained on the graph at some point. While a vertical stretch would only impact the y values, while the x values would remain the same. (1 vote) Landen J. Knapp 9 months agoPosted 9 months ago. Direct link to Landen J. Knapp's post “Okay this makes sense, bu...” Okay this makes sense, but in 5 or so minutes, I am bound to forget everything he just said. Its confusing. I need an easier way to remember this. • (4 votes) Heinz Doofensmirtz 9 months agoPosted 9 months ago. Direct link to Heinz Doofensmirtz's post “Just take some time to re...” Just take some time to review and think about what he discussed and work on gaining an intuitive understanding of it. (1 vote)Want to join the conversation?
Remember that x just represents an unknown number.
To find f(x) (you can think of f(x) as being y), you need to plug a number into x.
f(x)=|x|-3
x=-2
Plug -2 into x
|-2|-3
The absolute value of any number is positive. Thus, -2 will become 2. Then subtract. 2-3=-1.
When x=-2 y=-1
(-2, -1)
a is for vertical stretch/compression and reflecting across the x-axis.
b is for horizontal stretch/compression and reflecting across the y-axis. *It's 1/b because when a stretch or compression is in the brackets it uses the reciprocal aka one over that number.
h is the horizontal shift. *It's the opposite sign because it's in the brackets.
k is the vertical shift.
4:09
Do you normally do the opposite when going left to right?
What happens to the graph for f(x)=x when compared to the graph f(x)=x-5?
They could be shifted/translated, reflected, rotated, dilated, or compressed. So that's pretty much all you can do with a function, in terms of transformations. Hope that answered your question!
would the transformation of the problem be translation
Video transcript
So this red curve isthe graph of f of x. And this blue curve isthe graph of g of x. And I want to try to expressg of x in terms of f of x. And so let's seehow they're related. So we pick any x. And we could start righthere at the vertex of f of x. And we see that, at leastat that point, g of x is exactly 1 higher than that. So g of 2-- I couldwrite this down-- g of 2 is equal to f of 2 plus 1. Let's see if that'strue for any x. So then we can justsample over here. Let's see, f of 4is right over here. g of 4 is one more than that. f of 6 is right here. g of 6 is 1 more than that. So it looks like if we pickany point over here-- even though there's a little bitof an optical illusion-- it looks like theyget closer together. They do if you looktry to find the closest distance between the two. But if you look atvertical distance you see that itstays a constant 1. So we can actuallygeneralize this. This is true forany x. g of x is equal to f of x isequal to f of x plus 1. Let's do a few moreexamples of this. So right over here, hereis f of x in red again, and here is g of x. And so let's say we pickedx equals negative 4. This is f of negative 4. And we see g of negative4 is 2 less than that. And we see whatever f ofx is, g of x-- no matter what x we pick-- g of xseems to be exactly 2 less. g of x is exactly 2 less. So in this case, verysimilar to the other one, g of x is going tobe equal to f of x. But instead ofadding, we're going to subtract 2 from fof x. f of x minus 2. Let's do a few more examples. So here we have fof x in red again. I'll label it. f of x. And here is g of x. So let's think aboutit a little bit. Let's pick anarbitrary point here. Let's say we have in red here,this point right over there is the value of f of negative 3. This is negative 3. This is the pointnegative 3, f of 3. Now g hits that same valuewhen x is equal to negative 1. So let's think about this. g of negative 1 is equalto f of negative 3. And we could do thatwith a bunch of points. We could see that g of 0, whichis right there-- let me do it in a color you cansee-- g of 0 is equivalent to f of negative 2. So let me write that down. g of 0 is equal tof of negative 2. We could keep doing that. We could say g of 1,which is right over here. This is 1. g of 1 is equal tof of negative 1. g of 1 is equal tof of negative 1. So I think you seethe pattern here. g of whatever is equal to thefunction evaluated at 2 less than whatever is here. So we could say that g ofx is equal to f of-- well it's going to be 2 less than x. So f of x minus 2. So this is the relationship. g of x is equalto f of x minus 2. And it's importantto realize here. When I get f of x minus 2 here--and remember the function is being evaluated, this is theinput. x minus 2 is the input. When I subtract the 2, thisis shifting the function to the right, which is alittle bit counter-intuitive unless you go through thisexercise right over here. So g of x is equalto f of x minus 2. If it was f of x plus 2 wewould have actually shifted f to the left. Now let's think about this one. This one seems kind of wacky. So first of all,g of x, it almost looks like a mirrorimage but it looks like it's been flattened out. So let's think of it this way. Let's take the mirrorimage of what g of x is. So I'm going to try my best totake the mirror image of it. So let's see... It gets to about2 there, then it gets pretty close to1 right over there. And then it gets aboutright over there. So if I were to takeits mirror image, it looks something like this. Its mirror image if I were toreflect it across the x-axis. It looks something like this. So this right overhere we would call-- so if this is g of x,when we flip it that way, this is the negative g of x. When x equals 4, g ofx looks like it's about negative 3 and 1/2. You take the negative ofthat, you get positive. I guess it shouldbe closer to here-- You get positive3 and 1/2 if you were to take theexact mirror image. So that's negative g of x. But that still doesn't get us. It looks like weactually have to triple this value for any point. And you see it here. This gets to 2, butwe need to get to 6. This gets to 1, butwe need to get to 3. So it looks like thisred graph right over here is 3 times this graph. So this is 3 timesnegative g of x, which is equal tonegative 3 g of x. So here we have f of x is equalto negative 3 times g of x. And if we wanted to solve forg of x, right-- g of x in terms of f of x-- we wouldwrite, dividing both sides by negative 3, g of x isequal to negative 1/3 f of x.
