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## Algebra 2

### Course: 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 ?
• 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
• How do I know if I should reflex over the x-axis or y-axis?
(1 vote)
• -f(x) = reflection over x-axis
f(-x) = reflection over y-axis
-f(-x) = reflection over both axes
• at , why does Sal shift two to the left instead of two to the right?
• let f(x) = x^(1/3)
and g(x) = f(x+2)

f(x) evaluates to 0 when x=0. So the coordinates are x=0, y=0.
How do we make g(x) evaluate to 0? If x =-2, -2+2 = 0, so (-2+2)^(1/3) = 0. The coordinates are therefore x=-2, y=0

We can see that this is a shift to the left, rather than the right as you might initially expect.
• so...
at ,Sal said,"at x=-9, instead of getting 3, you should get 6"
i think he meant to say"x=-10"
i know this is probably very obvious, but i just thought i should post it
I hope it helps somebody!
• If you think about the square root function, √9=3 which is a nice number. √10 is an irrational number which could not be at a nice location on the graph.
• I'm not trying to be rude, but how in the world do you square a negative number?
• Well, for a number (-x), the square would be (-x)^2. This is just (-1)^2 * (x)^2, which is just x^2
• 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.
• 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.
• 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.