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## Algebra 2

### Unit 9: Lesson 6

Graphs of square and cube root functions

# 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.

## Want to join the conversation?

• Um, isn't the square root of -4 = sqr(4) * i? (i^2 = -1)? Shouldn't the function y = sqr(-x) be complex? • 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!)
• What's the order of operation when dealing with functions ? • How do I know if I should reflex over the x-axis or y-axis? • at , why does Sal shift two to the left instead of two to the right?
(1 vote) • Whats 3Sqrt of -x+4
(1 vote) • 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)
• 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) • 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)
• can someone explain transformation of functions widely?
(1 vote) • Can anyone try to explain the video a little bit better please?
(1 vote) • 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.

(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)
• 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) • 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)
• how would you solve if it is y = 4 square root x + 10?
(1 vote) • 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.