The Square Root Function
A square root function (or radical function) has a base equation which can be
written in the form:
y  x or f 
x  x
where Domain of the function is
x 0, x  , and the range is y 0, y 
There graph has the property that it starts at a fixed anchor point of (0, 0), and
extends in only one direction.
Example
a) By using transformations on the square root function
of the function
y  x , sketch the graph
y  x 3 2 .
b) State the domain and range of this transformed function.
Solution:
a) This graph has the anchor point (0,0) now at (-3,-2), thus it translates the base
function 3 units to the left and 2 units down.
b)
D f x x 3, x 
R f y y 2, y 
Example
a) By using transformations on the square root function
of the function
y 1 
y  x , sketch the graph
4x 8
.
16
b) State the domain and range of this transformed function.
Solution:
a) First we must rearrange this into the proper form.
y 1 
4 x 8
16

4x 8
1
16

4
x 2
1
16

1
x 21
4
1
 x 2 1
2
y a x p q
Pull out the
becomes
1
from within the radical sign (it
4
1
), so we only have a vertical stretch to
2
worry about.
This represents a reflection across the x-axis, a stretch of
1
, a horizontal shift of 2
2
units to the left, and a vertical shift of 1 unit up.
 So draw the anchor point at (-2, 1)
Go down one-half of the
square root of the number of
squares you went right with.
The negative sign in front of
radical sign means go down
 Then the next point is right 1 down
1
1
1 
2
2
1
 4 1
2
1
3
 The next point is right 9 down 9 
2
2
 The next point is right 4 down
Go right number which the square
root gives you an integer
b)
D f x x 2, x 
Use the graph to help you
with the range.
R f y y 1, y 
Example
Sketch
y
x 4
2
Solution:
We have two ways of doing this quickly
Method One: rewrite by pulling out the 2 in the denominator from the radical sign.
y
x 4
2
1
x 4
2
1
 
x 4
2

This is just a root function with anchor point at
a factor of
4,0 and an ugly vertical stretch by
1
0.707106...
2
We can do it but we will need our calculator to assist us.
Method Two: Leave the
1
1
in the radical and remember the
represent a horizontal
2
2
stretch by 2.
 Place our anchor point at
2 because you
stretch x-value by 2
4,0
Up the square root of x value
1 1
 Then right 2x4=8 up 4 2
 Then right 2x9=18 up 9 3
 Now right 2x1=2 up
Again, go right an amount where
the square root is an integer
Example
Sketch and state the domain and range for
y 3 2 8 4 x
Solution:
First place root function into proper form
y 3 2 8 4 x
2 8 4 x 3
2 4 
2 x 3
2 
2 2 x 3
4 
x 2 3
Since negative
signs inside and
outside of radical
sign
This function has anchor at (2, 3), and is reflected across both x and y axis, with a
final vertical stretch of 4.
For drawing purposes:
Place anchor at
2,3
Since negative
inside radical
Since negative
outside radical
Now from the anchor to the left 1 down
4  1 4
4  4 8
Next left 9 down 4  9 12
Next left 4 down
Now it is easy to find the domain and range by looking at the graph
D x x 2, x 
R y y 3, y 
But, how about find domain and range by looking at the original function:
positive
y 3 2 8 4 x
For the Domain the material inside the radical sign must be 0
That is,
8 4 x 0
8 4 x
2 x
x 2
From here the domain is
D x x 2, x 
Since negative sign is in
front of the radical sign
For the Range, we note that the graph goes down, therefore the range is below the yvalue of the anchor point. That is,
R y y 3, y 