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### Course: Algebra 1 (Eureka Math/EngageNY) > Unit 4

Lesson 14: Topic C: Lessons 20-22: Scaling and transforming graphs- Scaling & reflecting parabolas
- Scale & reflect parabolas
- Identifying function transformations
- Identifying horizontal squash from graph
- Reflecting & compressing functions
- Identify function transformations
- Transforming the square-root function
- Graphs of square-root functions
- Square-root functions & their graphs
- Radical functions & their graphs
- Graphs of square and cube root functions

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# Transforming the square-root function

Sal shows various examples of functions and their graphs that are a result of shifting and/or flipping y=√x. Created by Sal Khan.

## Want to join the conversation?

- What if it 's like 2x+3? with that radical sign over it ? How would you graph that?(4 votes)
- If you have the function y = sqrt(2x+3), you can rewrite the right hand side of this as:

y = sqrt(2(x+3/2)).

Then, using the property that sqrt(ab) = sqrt(a)*sqrt(b), you can rewrite this again as:

y = sqrt(2) * sqrt(x+3/2).

Now, notice that sqrt(2) is no more than a constant, you all you've done is stretched the graph vertically byu a factor of sqrt(2). Then, notice that under the second radical sign, you've got a shift to the left by 3/2. To show how this process makes sense, try graphing both y = sqrt(2x+3) and y = sqrt(2) * sqrt(x+3/2). You should get the same thing.

To graph it, know what the graph of y = sqrt(x) looks like first (its a parabola on its side with only the top half). Then, notice that you've shifted the graph to the left by 3/2 and stretched the entire graph by sqrt(2). All done!(13 votes)

- Why am I not getting any of this.(6 votes)
- lol so it's not just me(2 votes)

- Sal writes sqrt - (x + 3)

I can see the principle of reflection this radical function along line of x = -3. However will this not yield complex numbers? Thank you(4 votes)- It will not yield imaginary numbers as long as "x" is chosen carefully. We can find exactly for which values of x no complex numbers result. We do this by finding the domain of the function:

ƒ(x) = √[-(x + 3)]

The radicand must be greater than or equal to 0 in order for the function to yield only real numbers:

-(x + 3) ≥ 0

-x - 3 ≥ 0

-x ≥ 3

x ≤ -3

Therefore, all values of "x" that produce real numbers in the aforementioned equation are all numbers lesser than or equal to -3. This is also why the graph of the function starts at -3 on the x-axis and keeps going in the negative direction, but does not go in the positive direction at all. Comment if you have questions.(6 votes)

- Another way to say why x is positive instead of negative, you could say that y=√x -> or x=y^2. Because you want to move x three spaces to the left, you would do x=y^2 -3. Then, to convert it back, it would become y^2=x+3, or y=√x+3. idk just another way to understand it I guess.(2 votes)
- Your logic is good up to here: x=y^2 -3

So go back, add 3

y^2 = x+3

Then take the square root of both sides (the entire side)

y = √(x+3)

You can't just do the square root of "x"(3 votes)

- You said you have to make sure that whatever is under the radical will equal 0 at the origin like if we want to shift 2 to left we +2 to make sure that the number under radical is zero,but i want to ask why is that necessary to make it 0.?(2 votes)
- Radicals are tricky things. To evaluate points on a radical function, you want to think about values for x that make the radicand (value under the radical) equal to a perfect square, like 0, but also 1, 4, 9, 16, etc.(2 votes)

- how do you compute √-(x+3) ? I thought you couldn't have negatives under the radical(2 votes)
- does anyone know a video or link that describes how to find the x intercept of a transformation of the function y = cube root of x^2?(1 vote)
- I don't know of a video or link, but if you express the cube root of x^2 as x^(2/3) power, I'm sure you can find the x-intercepts by substituting 0 for y and solving for x with exponent rules.(3 votes)

- What would the graph of y = 3* sqrt(x) or y = sqrt(3x) or y = -3*sqrt(x), etc.?(1 vote)
- y=3*sqrt(x) would be the a slightly steeper version of y=sqrt(x). Think of it this way: each value for sqrt(x) is multiplied by three, making it it much steeper.

y=-3*sqrt(x) would be the exact mirror along the y axis of y=3*sqrt(x).

Hope this helped.(2 votes)

- At5:26, what is y when sqrt(x+3) is negative? I don't understand!(1 vote)
- sqrt(x) means positive (principal) sqrt always. If someone wants the negative sqrt they must write -sqrt(x). This is a rule.

When -(x+3) is negative (that is, to the right of -3), y = sqrt(-(x+3)) is undefined. At -3, -(x+3) is 0. As x moves to the left of -3, -(x+3) increases from 0, and so its sqrt, y, increases.(1 vote)

- When you graph a radical function how do you tell whether the x-value is negative or positive? I get that the y-value maintains it's negativity/positivity from the equation but the x seems to switch at random.(1 vote)
- All the possible x values come from the domain of the function. It's not random.(1 vote)

## Video transcript

So let's think
about the graph of y is equal to the
principal root of x. And then we'll start
playing around with this and see what happens
to the graph. So y is equal to the
principal root of x. Well, this is going
to be undefined if we want to deal
with real numbers. For x being any negative value. So the domain here is really x
is greater than or equal to 0. When x is 0, y is
going to be equal to 0. When x is 1, the principal
root of 1 is positive 1. So it's going to be like that. When x is 4, the
principal root of 4 is 2. When x is 9, the
principal root of 9 is 3. So this is what it is
going to look like. Which is going to
look more like this. So it's going to look
something like that. That's my best attempt
at actually graphing it. Now let's think
about what happens if we wanted to
shift it in some way. So let's say we
wanted to shift it up. Let's say we wanted
to shift it up by 4. So how would we do that? Well, whatever value
we're getting here, we want y to be 4 higher. So we could just add 4 to it. So we could just use y is equal
to the square root of x plus 4. So that would be like taking
this graph right over here. So let me copy and
then let me paste it. So it's like taking
this graph and we're shifting it up 1, 2, 3, 4. It would look that. Well, that was pretty
straightforward. Let me do it in
that same blue color so that you
recognize that that's that one right over there. But what if we wanted to shift
it, let's say, to the left. Let's say we wanted to
do something like this. Let's say we wanted to
shift it to the left by 3. Like that. So how would we define
the function then? And I'll do this in
this orange color. We want to shift by 3. So you're shifting
to the left by 3. So think about it. This point, y equaled
0, right over here, where whatever you put under
the radical was equal to 0. So what do you have to put
under the radical here to get 0? Well, here x is negative 3. So if you put x plus
3 under the radical then you are going to get--
and you take the square root of that-- you're going to get 0. So this right over here, this
orange function, that is y. Let me do it over here. y
is equal to the square root of x plus 3. And once again it might
be counter-intuitive. We went from square root of
x to square root of x plus 3. When we added 4 outside of the
radical that shifted it up. But when we add 3
inside of the radical instead of it shifting
it to the right, instead of shifting it that
way, it shifted it to the left. It made this point go
from 0 to negative 3. And the important
thing to realize is what makes y equal to 0. Over here y equals
0 when x is 0. Over here and over
here, y is equal to 0 when x plus 3 is equal to 0,
or x is equal to negative 3. And you could do
that for other points to see that it does
definitely shift to the left. And that's an important
thing to realize. This isn't actually just
for radical functions. This is actually for
functions in general. If you're just going to
add a 4-- add a number out here-- whatever you add is
going to shift it up or down. If this was a minus 4 it
would have shifted it down. But within, when you
replace the x with an x plus 3, when you replace
it with an x plus 3, this actually shifted
it to the left by 3. If you wanted to shift
it to the right by 3 you would put an x
minus 3 over there. Well, that's all interesting. But let's say that I wanted
to flip this thing over. So I wanted it to
look like this. Let me see if I can draw it. I want this graph to
look something like that. I'll try my best to-- to
look something like that. So it's flipped around. I could do a better
job than that. So we have that point on it. And then we're going
to go 3 and then we're going to have
that point on it. So I want it to
look-- actually I did a decent job the
first time I drew it-- I want to look
something like that. So essentially I want
to take its mirror image around the line x is equal to 3. How could I do that? Well, now my domain
is different. Now my domain-- it should
be undefined for anything where x is greater
than negative 3. And it should be defined
for any x that is less than or equal to negative 3. Or another way to
think about it is we need to flip the
sign of whatever we have under the radical. So this thing over here--
let me actually scroll over a little bit-- this
thing over here in green could be y is equal
to the square root of the negative of x plus 3. And I encourage you to try some
x values here to try it out. What we've done is
we've essentially flipped what happens
under the radical. Now in order to get a positive
value under the radical, now x plus 3 has to be negative. And the only way that
x plus 3 is negative, or the only way
that x plus 3 is, I guess you can
say nonpositive, is if x is less than negative 3. Now what if we wanted to
do something even more interesting? What if we wanted to flip
this one right over here over the horizontal
axis, over y equals 0? Well, then we're just
flipping what root we take. So that would be y is equal
to the negative square root of negative x plus 3. So it would look like this. It would look like that. And if we wanted to
shift that thing, we could just add or
subtract something outside of the radical. So let's say we wanted to do--
let me copy and paste this. So let's copy and
then let's see. Let's say that we wanted
to shift it down here. So instead of beginning at y
equals 0 right over here, where y equals negative 4,
then we would just subtract 4 outside
of the radical. So this thing and this thing--
I'm running out of colors here-- this thing
right over here would be y is equal to negative
square root of negative x plus 3 minus 4. And so we could keep
going on and on and on. But hopefully this
gives you a sense of the different ways you
could manipulate this thing. And we'll do some
more examples to get a better understanding of it.