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Graphing square and cube root functions

We can graph various square root and cube root functions by thinking of them as transformations of the parent graphs y=√x and y=∛x.

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  • male robot hal style avatar for user Smg88
    Um, isn't the square root of -4 = sqr(4) * i? (i^2 = -1)? Shouldn't the function y = sqr(-x) be complex?
    (12 votes)
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    • mr pants teal style avatar for user VincentTheFrugal
      Good catch! It is!.. Sorta! To get around this, mathematicians have defined the square-root function to only be defined for x >= 0, to avoid the very thing you noticed! Mathematicians sometimes change the definition of things if the definition causes situations that are more annoying than practical (yes, mathematicians actually do try to make things practical!)
      (12 votes)
  • male robot johnny style avatar for user Mohamed Ibrahim
    What's the order of operation when dealing with functions ?
    (3 votes)
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    • piceratops tree style avatar for user Carina Kaplan
      From the "show answers" of Khan Academy practice problems I've seen so far, you first translate right or left (h), then multiply the y-coordinates by whatever is in front of the radical (shrink if a < 1 or stretch if a > 1, and, if a is negative, flip over the x-axis), and then translate up or down (k)
      FYI: y=a√x-h +k
      (4 votes)
  • blobby green style avatar for user Daniel
    How do I know if I should reflex over the x-axis or y-axis?
    (0 votes)
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  • male robot hal style avatar for user kellanvega
    at , why does Sal shift two to the left instead of two to the right?
    (1 vote)
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  • blobby green style avatar for user Jamin Murray
    Whats 3Sqrt of -x+4
    (1 vote)
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    • mr pink green style avatar for user David Severin
      I assume you mean y=3√(-x+4). So we have to adjust it slightly to figure out the changes, y=3√(-1(x-4)). So three things are happening from y=√x. We have a scale factor of 3, a reflection across y, and a shift 4 units to the right. So y=√x gives points (0,0)(1,1)(4,2)(9,3) etc. start with shift to the right where these points become (4,0)(5,1)(8,2)(13,3). Reflect across y=4 to get (4,0)(3,1)(0,2)(-5,3). Finally, scale factor by 3 to get to points (4,0)(3,3)(0,6)(-5,9). You could have done these in different order.
      (1 vote)
  • blobby green style avatar for user Daniel
    I was doing a problem, it was the square root of -x+4. The hint told me that it would be easier to turn the square root of -x+4 to the square root of -(x-4). Why is it work like that? Shouldn't the starting point still be (0,-4)?
    (1 vote)
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    • mr pink green style avatar for user David Severin
      If you think about it, we first note that putting a negative in front of the x inside the √ sign causes a reflection across a y= line, so your question is about where it starts and goes to negative infinity. You can try numbers to see if it makes sense. So lets start at (0,-4) y = √(-x+4), so -4 gives √(-(-4)+4=√8, this ends up with 0 = √8 which is false and shows this point is not on the graph of y=√(-x+4). Try 0 for x and you get √4 = 2, so y has a value of 2 when x=0, already to the right of your supposition. Finally, if we try x = 4, you get √(-4+4)=√(0)=0, so you have the point (4,0). Just like other functions, the general transformation formula for square root would be y = a√(b(x-c))+d. So if you have √-(x-4) you see that c=4. The c value is such that a positive in the equation moves left and a negative moves right.
      (1 vote)
  • blobby green style avatar for user Sunnatbek Esonov
    can someone explain transformation of functions widely?
    (1 vote)
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  • stelly yellow style avatar for user micayla
    Can anyone try to explain the video a little bit better please?
    (1 vote)
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    • hopper cool style avatar for user Iron Programming
      Howdy micayla,

      Visualize a squared function in your head (y=x^2), but only in the first quadrant. Notice that if we want to make x the independent variable, we can easily do so by taking the square root of both sides (x=sqrt(y)). Now, the graph will look the same as before, but x is the independent variable. But, if we now swap variables, we will get the same graph as before but now the graph will oriented differently. Try graphing y=sqrt(x) on Desmos to see how this looks like.

      Now, there are several keys features we should notice about this function.
      (1.) There are no horizontal asymptotes. Although the graph does not increase very fast, as x gets larger y will continue to get larger forever. To prove this, take any infinitely larger y-value, and I can get you an x-value that is equal to that valued squared.
      (2.) This square root function will only be defined for x>=0, unless we are dealing with imaginary numbers (negative numbers under the square roots).
      (3.) Thus to draw the function, if we have the general picture of the graph in our head, all we need to know is the x-y coordinates of a couple squares (such as (2, 4)) and then we can graph the function, connecting the dots.

      Graphing functions by hand is usually not a super precise task, but it helps you understand the important features of the graph. This skill is especially important in Calculus, when we pay more attention to the maxima and minima of the function.

      Hope this helps.
      (1 vote)
  • piceratops ultimate style avatar for user tadtheyounger
    When I hit show answer for something like f(x) = sqrt(-x+4) it explains it like this: factor -1 out to get -1(x-4). Then multiply the -1 by that and it reflects over line x=4… why does it reflect over that specific line instead of something like the y axis. I have noticed a pattern with things like -x+b —> reflects over line x=b. However, I want to know why. Can someone please help?
    (1 vote)
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    • mr pink green style avatar for user David Severin
      I think it really matters which order you do the transformations in, since they are independent of each other, the order you do it in should not matter. You could reflect it across the x=0 line and then shift it 4 units to the right. The other choice is to shift it 4 units to the right first which would then require a reflection over x=4. When you shift a function along some vector <x,y>, what happens is that the shift would "act like" the new starting point is (0,0) even though it really is not. Thus, the shift from (0,0) to (4,0) would "act" like this is (0,0) and the reflection would then be across x=4. Do not know if this satisfies your curiosity.
      (1 vote)
  • blobby green style avatar for user thejeff
    how would you solve if it is y = 4 square root x + 10?
    (1 vote)
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    • piceratops ultimate style avatar for user Hecretary Bird
      You could graph this by looking at how it transforms the parent function of y = sqrt(x). y = 4sqrt(x) + 10 stretches the function vertically by a factor of 4, and translates it up by 10. Since the normal "vertex" of a square root function is (0,0), the new vertex would be (0,(0*4 + 10)), or (0,10). You can then just start at (0,10) and draw a wider curve than normal or plug in some points into the equation to help you draw the function more accurately.
      (0 votes)

Video transcript

- [Instructor] We're told the graph of y is equal to square root of x is shown below, fair enough. Which of the following is the graph of y is equal to two times the square root of negative x minus one? And they give us some choices here, and so I encourage you to pause this video and try to figure it out on your own before we work through this together. All right, now let's work through this together, and the way that I'm going to do it is I'm actually going to try to draw what the graph of two times the square root of negative x minus one should look like, and then I'll just look at which of the choices is closest to what I drew. And the way that I'm going to do that is I'm going to do it step by step, so we already see what y equals the square root of x looks like, but let's say we just want to build up. So let's say we want to now figure out what is the graph of y is equal to the square root of? Instead of an x under the radical sign, let me put a negative x under the radical sign. What would that do to it? Well, whatever was happening at a certain value of x will now happen at the negative of that value of x. So the square root of x is not defined for negative numbers. Now this one won't be defined for positive numbers. And the behavior that you saw at x equals two, you would now see at x equals negative two. The behavior that you saw at x equals four, you will now see at x equals negative four, and so on and so forth. So the y equals the square root of negative x is going to look like this. You've essentially flipped it over the y. We have flipped it over the y axis. All right, so we've done this part. Now let's scale that. Now let's multiply that by two. So what would y is equal to two times the square root of negative x look like? Well, it would look like this red curve, but at any given x value, we're gonna get twice as high. So at x equals negative four, instead of getting to two, we're now going to get to four. At x equals negative nine, instead of getting to three, we are now going to get to six. Now at x equals zero, we're still going to be at zero 'cause two times zero is zero, so it's going to look, it's going to look like that. Something like that, so that's y equals two times the square root of negative x. And then last but not least, what will y, let me do that in a different color. What will y equals two times the square root of negative x minus one look like? Well, whatever y value we were getting before, we're now just going to shift everything down by one. So if we were at six before, we're going to be at five now. If we were at four before, we're now going to be at three. If we were at zero before, we're now going to be at negative one, and so our curve is going to look something like, something like that. So let's look for, let's see which choices match that. So let me scroll down here, and both C and D kind of look right, but notice, right at zero, we want it to be at negative one, so D is exactly what we had drawn. And at nine, we're at five. Or at negative nine, we're at five. At negative four, we're at three, and at zero, we're at negative one. Exactly what we had drawn. Let's do another example. So here, this is a similar question. Now they graphed the cube root of x. Y is equal to the cube root of x, and then they say which of the following is the graph of this business? And they give us choices again, so once again, pause this video and try to work it out on your own before we do this together. All right, now let's work on this together and I'm gonna do the same technique. I'm just gonna build it up piece by piece. So this is already y is equal to the cube root of x. So now let's build up on that. Let's say we want to now have an x plus two under the radical sign. So let's graph y is equal to the cube root of x plus two. Well, what this does is it shifts the curve two to the left. And we've gone over this in multiple videos before, so we are now here, and you could even try some values out to verify that. At x equals zero, at x equals zero, or actually, let me put it this way. At x equals negative two, you're gonna kick the cube root of zero, which is right over there. So we have now shifted two to the left to look something, to look something like this, and now, let's build up on that. Let's multiply this times a negative, so y is equal to the negative of the cube root of x plus two. What would that look like? Well if you multiply your whole expression, or in this case, the whole graph or the whole function by a negative, you're gonna flip it over the horizontal axis. And so it is now going to look like this. Whatever y value we're gonna get before for a given x, you're now getting the opposite, the negative of it. So it's going to look, it's going to look like that, something like that. So that is y equal to the negative of the cube root of x plus two. And then last, but not least, we are going to think about, and I'm searching for an appropriate color. I haven't used orange yet. Y is equal to the negative of the cube root of x plus two, and I'm going to add five. So all that's going to do is take this last graph and shift it up by five. Whatever y value I was going to get before, now I'm going to get five higher. So five higher, let's see. I was at zero here, so I'm now going to be at five here. So that's going to look, it's going to look something, something like, something like that. And I'm not drawing it perfectly, but you get the general, the general idea, now let's look at the choices. And I think the key point to look at is this point right over here, that in our original graph, was at zero, zero. Now it is going to be at negative two, comma, five. So let's look for it, and it also should be flipped. So on the left hand side, we have the top part and on the right hand side, we have the part that goes lower. So let's see. So A, C, and B all have the left hand side as the higher part and then the right hand side being the lower part, but we wanted this point to be at negative two, comma, five. A doesn't have it there. B doesn't have it there. D we already said goes to the wrong direction. It's increasing. So let's see, negative two, comma, five, it's indeed what we expected. This is pretty close to what we had drawn on our own, so choice C.