Aaron Thomas
Jacob Wefel
Tyler Sneen
By the end of this lesson we will introduce the
terminology that is used to describe functions
These include: Domain, Range, Continuity,
Discontinuity, upper and lower bound, Local
and absolute maximums and minimums, and
asymptotes
The domain of a function is all of the possible
x-values the function can have. It can be
expressed as an inequality
The Range of a function is all of the possible
y-values the function can have. It is also
expressed as an inequality
Domain: All Real Numbers
Range: All Real Numbers
Domain: x> -1
Range: x>-5
A graph has continuity if its graph is
connected to itself throughout infinity. There
are no asymptotes or holes in the graph
A Graph has removable discontinuity if its
graph has a hole where one x value was
removed from the domain
A graph has infinite discontinuity if its graph
has an asymptote that can not be replaced
with only one value
Jump Discontinuity
Removable Discontinuity
A function is bounded above or below if the
graph’s range doesn’t extend past a certain
point above or below.
A function is “Bounded” if the function’s
range doesn’t extend below or above certain
points
If the function has no restrictions on its
range’s extent the function is considered
“unbounded”
This sine function is bounded above and
below at 1 and -1
A Local Maximum/Minimum of a function is
the highest/lowest point of the range in the
surrounding window of the graph
The absolute maximum/minimum of a
function is the highest/lowest point of the
entire range of the graph
Local Min: 3, -4, 4
Local Max: 5
Absolute Max: None (Graph goes infinitely
upward)
Absolute Min: -4
A horizontal asymptote is a part of the
function which gets infinitely close to a Yvalue but never touches it
A Vertical asymptote is a part of the function
which gets infinitely close to a x- value but
never touches it
Identify any horizontal or vertical asymptotes of
the graph of
You would first start by foiling the
denominator… = (x+1)(x-2)
This means that the graph has vertical
asymptotes of x=-1 and x=2
Because the denominator’s power is bigger than
the numerator’s, y = 0 no matter what the value
of x is
Now you have x/((x+1)(x-2)) = 0
This means that the horizontal asymptote is
zero
0
