ABSOLUTE VALUE AND
RECIPROCAL FUNCTIONS
Chapter 7
EXAMPLE
How many triangles are in the diagram below?
7.1 – ABSOLUTE VALUE
Chapter 7
ABSOLUTE VALUE
For a real number a, the absolute value is written as |a| and is a positive number. For
example:
|5| = 5
|–5| = 5
Absolute value can be used to represent the distance of a number from zero on a realnumber line.
Evaluate:
|3|
|–7|
EXAMPLE
Evaluate the following:
a) |4| – |–6|
b) 5 – 3|2 – 7|
a) |4| – |–6| = 4 – 6 = –2
b) 5 – 3|2 – 7| = 5 – 3|–5| = 5 – 3(5) = 5 – 15 = –10
c) Try it!
c) |–2(5 – 7)2 + 6|
PG. 363-367 #1, 6,
7(A,C,E), 11, 16
Independent
Practice
HANDOUT
Answer the questions on the
“Investigating Absolute Value Functions”
worksheet to the best of your ability.
7.2 – ABSOLUTE VALUE
FUNCTIONS
Chapter 7
ABSOLUTE VALUE FUNCTIONS
For what values of x is the function y = |x|
equivalent to y = x?
 when x ≥ 0
When x < 0, what is the function represented
by
y = |x|?
 y = –x
We can write this as a piecewise function:
ì x, if x ³ 0
y=í
î-x, if x<0
EXAMPLE
Consider the absolute value function y = |2x – 3|.
a) Determine the y-intercept and the x-intercept.
b) Sketch the graph.
c) State the domain and range.
d) Express as a piecewise function.
a) The y-intercept is at x = 0.
y = |2x – 3|
y = |2(0) – 3|
y = |–3|
y=3
The y-intercept is (0, 3).
The x-intercept is at y = 0.
0 = |2x – 3|
0 = 2x – 3
x = 3/2
The x-intercept is at (3/2, 0).
b)
x
y
-1
5
0
3
3/2
0
3
3
4
5
EXAMPLE
Consider the absolute value function y = |2x – 3|.
a) Determine the y-intercept and the x-intercept.
b) Sketch the graph.
c) State the domain and range.
d) Express as a piecewise function.
x
b)
y
-1
5
0
3
3/2
0
3
3
4
5
The x-intercept is call an
invariant point because it’s a
part of both functions.
c) D: {x | x E R}
R: {y | y ≥ 0, y E R}
d) The equation on the
right is just y = 2x – 3.
3
ì
2x
3,
if
x
³
ïï
2
f (x) = í
ï-(2x - 3), if x < 3
ïî
2
What’s the one on the
left?
It’s just –(2x – 3)!
EXAMPLE
Consider the absolute value function f(x) = |–x2 + 2x + 8|.
a) Determine the y-intercept and the x-intercepts.
b) Sketch the graph.
c) State the domain and range.
d) Express as a piecewise function
a) The y-intercept is at x = 0.
b) What’s the vertex of the function
 f(0) = |–(0)2 + 2(0) + 8|
= |8|
=8
f(x) = –x2 + 2x + 8
The x-intercepts are when y = 0.
 0 = |–x2 + 2x + 8|
 0 = –x2 + 2x + 8
 0 = –(x – 4)(x + 2)
x=4
 x = –2
f(x) = –(x2 – 2x) + 8
 f(x) = –(x2 – 2x + 1 – 1 ) + 8
 f(x) = –(x2 – 2x + 1) + 1 + 8
 f(x) = –(x – 1)2 + 9
 The vertex is (1, 9)
Use your calculator, or complete the square:
EXAMPLE
Consider the absolute value function f(x) = |–x2 + 2x + 8|.
a) Determine the y-intercept and the x-intercepts.
b) Sketch the graph.
c) State the domain and range.
d) Express as a piecewise function
Recall:
 y-intercept is (0, 8)
 x-intercepts are (4, 0) and (–2, 0)
 Vertex is (1, 9)
c) D: {x | x E R}
R: {y | y ≥ 0, y E R}
d)
ìï-x 2 + 2x + 8, if - 2 £ x £ 4
f (x) = í
2
ïî-(-x + 2x + 8), if x < -2 or x > 4
PG. 375-379, #2, 5, 7,
10, 12-14.
Independent
Practice
7.3 - ABSOLUTE VALUE
EQUATIONS
Chapter 7
ABSOLUTE VALUE EQUATIONS
When solving equations that involve absolute value
equations you need to consider two cases:
Case 1: The expression inside the absolute value
symbol is positive or zero.
Case 2: The expression inside the absolute value
symbol is negative.
EXAMPLE
Solve: |x – 3| = 7
Consider the equation as a piecewise function:
ì x - 3, if x ³ 3
x-3 =í
î-(x - 3), if x < 3
Case 1:
Case 2:
x–3=7
 x = 10
–(x – 3) = 7
 x – 3 = –7
 x = –4
The solution is x = 10 or x = –4.
TRY IT
Solve |6 – x| = 2
EXAMPLE
Solve |2x – 5| = 5 – 3x
What is the x-intercept of y = 2x – 5?
 0 = 2x – 5
 5 = 2x
 x = 5/2
Consider:
ì
5
2x
5,
if
x
³
ïï
2
f (x) = í
ï-(2x - 5), if x < 5
ïî
2
Case 1: (x ≥ 5/2)
2x – 5 = 5 – 3x
 5x = 10
x=2
Case 2: (x < 5/2)
–(2x – 5) = 5 – 3x
 –2x + 5 = 5 – 3x
x=0
EXAMPLE
Solve: |3x – 4| + 12 = 9
 |3x – 4| = –3
Is there any possible way that the absolute value of something is equal to –3?
 No solution.
EXAMPLE
Solve: |x – 10| = x2 – 10x
PG. 389-391, #4, 5, 6,
9, 11, 22, 23
Independent
Practice
7.4 – RECIPROCAL
FUNCTIONS
Chapter 7
EXAMPLE
Sketch the graphs of y = f(x) and its reciprocal function y = 1/f(x), where f(x) = x.
Examine how the functions are related.
x
y=x
y = 1/x
–5
–5
–1/5
–2
–2
–1/2
–1
–1
–1
–1/2
–1/2
–2
–1/10
–1/10
–10
0
0
Undef.
1/10
1/10
10
1/2
1/2
2
1
1
1
2
2
1/2
RECIPROCAL FUNCTIONS
An asymptote is a line whose
distance from a curve approaches
zero.
This graph has two pieces, that both
approach the vertical asymptote,
which is defined by the nonpermissible value of domain of the
function, and a horizontal asymptote,
defined by the value that is not in the
range of the function.
What is the vertical asymptote?
What is the horizontal asymptote?
EXAMPLE
Consider f(x) = 2x + 5.
a) Determine its reciprocal function y = 1/f(x).
b) Determine the equation of the vertical asymptote of the reciprocal function.
c) Graph the function y = f(x) and its reciprocal function y = 1/f(x).
a) The reciprocal function is:
1
y=
2x + 5
b) The vertical asymptote is always
the non-permissible values of the
function.
2x + 5 = 0
 2x = –5
 x = –5/2
There is a vertical asymptote at
x = –5/2
c)
Characteristic
f(x) = 2x +5
f(x)=1/(2x + 5)
xintercept/asy
mptotes
x-intercept at
x = –5/2
Asymptote at
x = –5/2
Invariant
points
2x + 5 = 1
 x = –2
 (–2, 1)
(–2,1)
Invariant points 2x + 5 = –1
are at y = 1 and  x = –3
y = –1.
 (–3,–1)
(–3,–1)
EXAMPLE
Consider f(x) = 2x + 5.
a) Determine its reciprocal function y = 1/f(x).
b) Determine the equation of the vertical asymptote of the reciprocal function.
c) Graph the function y = f(x) and its reciprocal function y = 1/f(x).
Characteristic
f(x) = 2x +5
f(x)=1/(2x + 5)
xintercept/asy
mptotes
x-intercept at
x = –5/2
Asymptote at
x = –5/2
Invariant
points
2x + 5 = 1
 x = –2
 (–2, 1)
(–2,1)
2x + 5 = –1
 x = –3
 (–3,–1)
(–3,–1)
EXAMPLE
Consider f(x) = x2 – 4.
a) What is the reciprocal function of f(x)?
b) State the non-permissible values of x and the equation(s) of the vertical asymptote(s)
of the reciprocal function.
c) What are the x-intercepts and y-intercepts of the reciprocal function?
d) Graph the functions.
a)
y=
1
x2 - 4
b) What are the non-permissible values?
 x2 – 4 = 0
 (x – 2)(x + 2) = 0
x=2
 x = –2
 The vertical asymptotes are at x = ±2
c) How can I find the x-intercept of the
the reciprocal?
1
0
=
 Let f(x) = 0
2
x -4
 There is no solution, so there is no
x-intercept.
y-intercept is y = –1/4
y-intercept:
f (x) =
1
1
Þ
f
(x)
=
4
02 - 4
PG. 403-408, #3, 5, 7,
8, 9, 10, 12
Independent
practice