Function V
Y-axis
Y-axis
y  6  2x
f (x)  x 2  4x
6
(2, 4)
√6

X-axis

X-axis
0
3
4
(2,-4)
y1 = 6 -2x
Square-root and Absolute Functions
By Mr Porter
Square-Root Function: y 
f (x)
The important feature of this function is that f(x) must be a ZERO or POSITVE value,
otherwise the function is undefined in the real number system. We will at this stage of
the course, restrict f(x) to be linear functions, such as f(x) = 2x + 3 or f(x) = 4 - x.

The function, g(x)  f (x) , is equivalent to half of a horizontal parabola and has
similar graphing features.
Example
 the function: , y  x  5 , squaring both sides
Consider
y2 = x – 5 , re-arrange to make x the subject.

x = y2 + 5 , which is the form of a horizontal parabola.
The important fact in this example is that: x – 5 ≥ 0, to have any real answer.
This means, the Domain is x ≥ 5 for all x in R,
and the Range is y ≥ 0 for all y in R.
Graphing : y 
To sketch y 
f (x)
f (x) , start with y = f(x).
Example 1: Sketch y  2x  5
Hence,
 state clearly the functions

domain
and range.
Sketch the
graph of y = 2x - 5.
Now, take the square-root, top half of a
parabola
Domain is found by solving 2x - 5 ≥ 0
Example 2: Sketch y  6  2x
Hence, state clearly the functions domain
and range.
Solution: Find
 domain first, solve 6 – 2x ≥ 0
-2x ≥ -6
x≤3
This mean the curve
goes to the left.
Sketch y1 = 6 - 2x on the number plane.
Now, take the square-root, top half of a
parabola left.
Y-axis
y1=2x-5
Y-axis
y  6  2x
y  2x  5
√6
X-axis
X-axis
2.5


-5
6
Remember, down here
y = 2x - 5 is negative!
Domain:
all x ≥ 2.5 in R
Range:
All y ≥ 0 in R
3
Domain:
all x ≤ 3 in R
Range:
All y ≥ 0 in R
y1 = 6 -2x
Harder Examples: Sketch the graph of the following function and clear state the
domain and range of each.
a) f (x)  2  x 1
b) f (x)  5  2x  4
Sol: Find domain first, solve x +1 ≥ 0
x +1 ≥ 0 ==> x ≥ -1 , curves to the right .

Sketch y1 = x + 1 on the same number plane
and y2 = 2.
Now, sketch the square-root curve to the
right, starting at x = -1. Slide the curve up 2!
y
Sol: Find domain first, solve 2x - 4 ≥ 0
2x – 4 ≥ 0 ==> x ≥ 2 , curves to the right .
Sketch y1 = 2x – 4 on the same number plane
and y2 = 5.
Because of the (–) in-front of the square-root,
the curve goes under. Slide the curve up 5!
y = x+1
y1 = 2x- 4
y2 = 5
f (x)  2  x 1
f (x)  5  2x  4
y=2
2

-1

x
Domain: All x ≥ -1 in R
Range: All y ≥ 2 in R
Domain: All x ≥ 2 in R
Range: All y ≤ 5 in R
Exercise: For each of the following:
a) Sketch the function
b) Define the domain and range of the function
a) f (x)  4  x
b) f (x)   x  6
c) f (x)  2x  6  2
y
y

2
y
x
-6
-3
5
x
x
4
Domain: all x ≤ 4 in R
Range: all y ≥ 0 in R
-√6
Domain: all x ≥ -6 in R
Range: all y ≤ 0 in R
-2
6 2

Domain: all x ≥ -3 in R
Range: all y ≥-2 in R
Absolute Value Functions. y = | f(x) |
The absolute symbols | .. | given the value as a POSITIVE number, as in distance.
Examples: Sketch and define domain and range.
a) f (x)  2x  3
b) f (x)  x 2  4x
Step 1: Sketch f(x) = 2x – 3
Step 1: Sketch f(x) = x2 - 4x
Step 2: at the x-intercept, reflect graph UP.
Step 2: at the x-intercepts, reflect negative part of graph UP.
f (x)  2x  3

y
f (x)  x 2  4x
3
(2, 4)
11/2
x
-3

0
4
(2,-4)
Domain: all x in R
Range: all y ≥ 0
Domain: all x in R
Range: all y ≥ 0
More Examples
2
c) f (x) 
x 1
Step 1: Sketch f(x) =
d) f (x) 
2
, asymptote at x = 1, y = 0
x 1
2
x 1
2
, asymptote at x = 1, y = 0
x 1
Step 2: Reflect the positive part of the graph DOWN
below the x-axis.
Step 1: Sketch f(x) =
Step 2: Reflect the negative part of the graph UP above the x-axis.



f (x) 
2
x 1
2
2
[An absolute graph can be NEGATIVE]

1
-2
1
f (x) 
-2
2
x 1

Domain: all x in R, x ≠ 1
Range: all y in R, y > 0
Domain: all x in R, x ≠ 1
Range: all y in R, y < 0
Exercise: For each function:
i) Sketch the Function
ii) Write down the domain and range.
b) f (x) 
a) f (x)  4  2x
1
2x  3
4

Domain: all x in R
Range: all y ≥ 0 in R
2
c) f (x)  x  4
2
1 12
Domain: all x in R, x ≠ 1.5
Range: all y ≥ 0 in R, y ≠ 0
1
3


Note: f(x) = |(x - 2)(x + 2) |
d) f (x)  x 2  4x 12
Note: f(x) = |(x - 6)(x + 2) |

4
(2,16)
-2
2
-4
Domain: all x in R
Range: all y ≥ 0 in R
-2
2
(2,-16)
6
Domain: all x in R
Range: all y ≥ 0 in R