Chapter Five
Rational Functions
A rational number is a number that is the quotient of two integers. A
rational function is the quotient of two polynomials.
Since creating a rational function involves division, for the first time you
must be careful about the domain you are using. Unlike addition,
subtraction, and multiplication, division of two numbers is not always
possible. Dividing by the number zero is an illegal operation and cannot be
done. Your calculator will declare a mathematical error if you attempt such
a division.
When you analyze a rational function, your first step must be to find out if
there are any input values that would make the value of the polynomial in
the denominator zero. You must reject such values from the domain to
prevent the function from attempting to perform an illegal operation.
Examples of rational numbers:
,
,
,6
,0
Irrational numbers:
,
,
,
Rational functions:
,
,
If you could divide by 0, then
1
0
= z would imply that z
which is impossible.
such quotient.
0 = 1,
There is no
To remind yourself not to use these values in the function, you usually
show them on the graph with dotted lines parallel to the y-axis passing
through the excluded numbers on the x-axis. Your graph cannot cross such
a dotted line since the intersection would indicate a value obtained by your
function using that x-value, which is impossible. Since you may use any
other x-value but these, you can use inputs as close to the eliminated ones
as you want. A line or curve which can be approached as closely as you
wish, although perhaps not actually reached, is called an asymptote. The
dotted lines indicating the values you have eliminated from your domain are
therefore called (vertical) asymptotes.
These asymptote lines have the effect of breaking up your domain into
separate pieces and you analyze the function over each piece to produce
separate parts of the graph. You use the same reasoning you have for
previous graphs to complete the rational function graphs. The graphs
obtained for rational functions by these methods are quite accurate.
Your basic method for producing the graph of a function has been to find
a starting point (often the intercept), then move through the domain from
your starting value as far as possible in each direction to analyze the endbehavior, and sketch the graph considering other important properties of the
function (like the number and location of turning-points).
Vertical asymptotes break up the
graph into several pieces.
82 CHAPTER 5: RATIONAL FUNCTIONS
For non-polynomial functions you should always check the domain first
before applying this method. You must be sure not to use values that are
not allowed or that you do not wish to use as inputs. Since all polynomial
functions (including linear and quadratic) have domains consisting of all
real numbers, checking the domain is not a crucial step in obtaining the
graphs for these functions, although you may limit your domain to
reasonable values for some applied problems. However, with the rational
functions, it is necessary for you to remove any impossible input values
from the domain before you start. You can then go on with the steps you
usually take.
Steps to obtain the graph of a rational function:
First remove from the domain any x-values which produce a value of
zero in the denominator. You can indicate these missing values with
vertical asymptote lines (often represented with dotted lines).
Next, for each piece of the graph, select a value to obtain a starting point
and plot this point.
Move through the domain in both directions from your starting value as
far as you can. This may be toward the ends of the domain at the
extremely large positive and negative x-values, or toward the values you
have removed, indicated by the vertical asymptotes. Analyze the
behavior of the function at the ends of each piece of domain and sketch
the curve representing this behavior. Note that the graph may be split up
into several pieces.
For an example in applying these steps, look at f1 (x) =
1
.
x
The first thing you must do is eliminate from the domain any x-values
that produce a denominator value of zero. This time, the only such
x-value is x = 0. The graph cannot cross the y-axis anywhere, and this
function has no intercept. The y-axis, x = 0, serves as a vertical
asymptote. This asymptote divides the domain into two separate
pieces, those x-values that are negative (to the left of x = 0) and those
that are positive (to the right of x = 0). Now analyze the function
separately in each piece of the domain.
Looking at the right-hand side first, you may choose an input value to
use to obtain a starting point. For example, you might select x = 2.
Then f1(2) =
. You may plot this point.
CHAPTER 5: RATIONAL FUNCTIONS
Now move through the domain from x = 2 to your left. You cannot go
as far as x = 0, since this value is not in the domain. However, you can
use x-values very close to x = 0, as long as they are all above it. As you
do this be aware of what is happening to the values obtained by the
function. The smaller (closer to x = 0) you make the input values, the
larger the values of f1(x) are. Dividing the number 1 by smaller and
smaller positive values produces larger and larger quotients. Your
graph should show this by curving upward as you move from your
( ) to the left (but not beyond the y-axis).
starting point 2,
1
2
Next, move from x = 2 to the right toward the larger positive x-values.
As the input values become higher positive numbers, the values of f1(x)
become smaller (closer to zero), since dividing the number 1 by large
positive numbers produces small positive quotients. When you move
further to the right, your input value becomes larger and your quotient
becomes closer to zero. Your graph approaches the x-axis (y = 0) from
above and moves closer and closer to it. Eventually the curve appears
to merge with the x-axis although the two never quite meet. The x-axis
acts as a horizontal asymptote to your curve since the graph gets as
close to the axis as you want if you move far enough to the right.
Note what happens as your x-values
move from x = 2 toward x = 0:
f1(2) =
f1(1) =
f1( ) =
=2
f1(.1) =
= 10
f1(.01) =
= 100
f1(.001) =
= 1000
As you use increasingly large
positive x-values, the values of f1(x)
get close to 0:
f1(2) =
f1(5) =
f1(10) =
f1(100) =
f1(1000) =
That takes care of the graph for the domain to the right of x = 0.
Now work on the domain to the left of x = 0. Again, you may select a
value to use to get a starting point in this part of the domain. You might
choose x = –2. Then your starting point on this side of the y-axis is
( −2, f1 (−2) ) = ( −2, − 12 ) .
Moving through domain values to the right of x = –2 (up to but not
including x = 0) produces f1(x) values that become more negative since
you are now dividing your numerator, 1, by negative numbers that are
becoming smaller (that is, in absolute value). This causes the graph to
drop steeply toward extremely negative values as you move further to
the right (but not past the y-axis).
83
Note the values of f1(x):
f1(–2) =
f1(–1) =
f1(–.1) =
f1(–.01) =
f1(–.001) =
= –10
= –100
= –1000
84 CHAPTER 5: RATIONAL FUNCTIONS
Finally, move through the domain from x = –2 toward increasingly
negative x-values (to the left). You obtain the values of f 1(x) by
dividing your numerator, 1, by negative numbers that are becoming
larger in size (further from x = 0). The quotients you obtain are all
negative, but are getting closer to 0. The graph of the function for this
part of the domain is approaching the x-axis as you move further to the
left, but on this side you come from below the axis since all the values
of the function are negative. The y-axis, x = 0, is a horizontal
asymptote on this side too.
This completes the graph of the function f1(x) =
.
You can also use the methods you previously learned to determine the
graphs of functions obtained from this one by stretching, translating and
reflecting.
For example, the graph of
is
stretched vertically compared to the
graph of . The values are three times
as large in both positive and negative
directions. The figure on the right
shows the graphs of
and
.
The graph of
that of
is the reflection of
across the x-axis. The
values of
are the same as those of
but with opposite signs. The graph
of
is given on the left.
The values of f1(x) now are:
f1(–2) =
f1(–5) =
f1(–10) =
f1(–100) =
f1(–1000) =
CHAPTER 5: RATIONAL FUNCTIONS
The graph of
is the translation of
the graph of
to the right by 4 units.
The values obtained by
are the
same as those obtained by
but for
85
Sketch the graphs of:
a.
, b.
, c.
–2
a.
x-values that are 4 units higher. The
graph of
is given at the right.
b.
The graph of
+ 2 is the translation of
the graph of
two units upward. Notice
that means the horizontal asymptote also
moves up 2 units from the x-axis to
y = 2. The graph is given on the left.
c.
For a more complex example, let f2(x) =
.
One way to approach this is to rewrite the expression for the function
=
=
=3+
In this form you can see that you can obtain the graph from the graph
of
by translating up 3 units, to the right 2 units, and a large vertical
stretch by a factor of 14. The horizontal asymptote is y = 3.
(The graph is given on page 88.)
If you do not want to carry out all the algebraic steps to rewrite the
expression in recognizable form, you can obtain the graph from the
original expression by the usual procedure.
First, you must exclude x = 2 from the domain because the
denominator, x – 2, would be zero for x = 2 and this is not permissible.
You show this exclusion from the domain with a dotted line through
x = 2 (the vertical asymptote). The graph is in two pieces.
You can rewrite the expression in the
numerator by looking for a multiple
of (x – 2), the denominator, so you
can divide. Since the coefficient of
the x-term in the numerator is
already 3, what you need to have is
3(x – 2) or 3x – 6. What you do
have is 3x + 8, which is 14 higher
than what you want. Adding 14
to 3x – 6 is just another way of
writing 3x + 8. You then can divide
3x – 6 by x – 2 for a quotient of 3,
and have a remainder of 14. This
method often works.
86 CHAPTER 5: RATIONAL FUNCTIONS
The intercept is at (0, –4) since f2(0) =
=
= –4. This gives
you a point for the left part of the graph.
Now, analyze the end-behavior. On the left-hand side, extremely
negative x-values produce large negative numbers in both numerator
and denominator, so the quotient is a positive number. However, you
can say something much more precise than that.
Remember that the x-value you are using is very large in size. The term
3x + 8 tells you to multiply this very negative number by 3, producing
a more negative value, then to add 8. However, adding 8 to an
extremely negative number does not have much effect, and the larger
in size the x-value is, the less significant the addition of 8. The
dominant term in the numerator is 3x.
Similarly, in the denominator, the size of the x-value is what is
important. Subtracting 2 from a number that is already very negative
has very little effect. The value of the term x dominates that of x – 2.
Specifically,
You can see that as the input values become more negative, the 8 added
in the numerator and the 2 subtracted in the denominator become less
and less significant. When x = –10, the –22 obtained in the numerator
is quite different from –30, but when x = –100, the –292 is pretty close
to the –300 you had before adding 8. For x = –1000, there is very little
difference between the calculated –2992 and the approximated –3000.
The same is true for the denominator, which has a value very close to
x when x is far to the left.
When the x-values are very large in size, the only terms that are
significant are those that have the highest degree, both in numerator and
denominator. In this example, the highest power of the variable is the
first power both in the numerator and in the denominator.
Therefore, for x-values that are very far to the left,
= 3.
CHAPTER 5: RATIONAL FUNCTIONS
The graph of this function on the extreme left is a curve that is very
close to y = 3. From the preceding calculations, you can also see that
the values of the function are all below 3, and get closer to 3 when you
use more negative values for inputs. The line y = 3 is a horizontal
asymptote for the graph. Remember that you found this by looking at
the values of the function for extremely negative x-values. Don't forget
that the function values are determined by the input values. Also note
that the asymptote is not part of the graph, but is a guide line to help
make the graph more accurate. The only parts of the asymptote line
that are helpful are the far ends that show what value the function is
approaching when the x-values are large in size.
At the other end of this piece of the graph you have numbers close to,
but less than x = 2. When you use these numbers in the function, the
numerator has a value near 3(2) + 8 = 14. The denominator has a value
near but slightly below 0. Remember the denominator could never be
equal to 0. Therefore, when you use values just under 2 for x, your
quotient is a negative number that is large in size (since you are
dividing by a number that is almost 0).
Specifically,
The left-hand part of the graph shows a curve that is very close
(although slightly below) f(x) = 3 for x-values on the extreme left,
passes through the intercept (0, –4), and then heads rapidly downward
when you get near the dotted line indicating the excluded value x = 2.
On the right side, you can get a starting point by using any value greater
than 2 for x. Find and plot such a point.
Next you see that x-values slightly above x = 2 produce numerators
near 14 divided by denominators that are slightly above 0. This gives
you quotients that are very large in size, and positive.
87
88 CHAPTER 5: RATIONAL FUNCTIONS
The graph shows a curve to the right of x = 2 that is very high near
x = 2, then drops and eventually passes through your starting point.
Finally, use the x-values on the extreme right of your domain, the very
large positive numbers. These numbers give very large positive
numbers in both numerator and denominator. Again, closer
examination shows that since you are using very large values for x, the
8 added in the numerator and the 2 subtracted in the denominator do
not play significant parts in the calculations.
As you use larger and larger values for x, the positive quotients get
closer to 3, and since the numerator is always more than 3 times the
denominator, these quotients are always larger than the limiting value
3.
On this side, the line y = 3 is again a (horizontal) asymptote and the
graph approaches this line (from above) as you move your x-values to
the right.
Sketch the graphs of:
a.
b.
a.
b.
Complete graph of f2(x).
CHAPTER 5: RATIONAL FUNCTIONS
A different sort of example is given by f3(x) =
.
There are now two x-values, x = –2 and x = 5, that make one of the
factors in the denominator equal to 0. You must remove both these
numbers from the domain to avoid having the denominator take on a
value of 0. The graph is in three pieces, one for the domain from the
extreme left up to x = –2, another between x = –2 and x = 5, and the
third from above x = 5 to the extreme right.
The intercept,
, gives you a starting point for the middle piece.
Then when you move through the domain from x = 0 toward x = –2, the
values of f3(x) are very large in size because the denominator is close
to 0. The numerator, –4, is a negative number. When you use x-values
very close to, but above, x = –2, the factor (x + 2) is close to 0, but
positive, while the factor (x – 5) is negative. The denominator, the
product of a positive and a negative number, is negative, and the
function values, calculated with a negative numerator and denominator
are positive. Therefore, the graph shows large positive values when the
x-values used are slightly above x = –2.
As you move through your domain toward x = 5, using x-values slightly
smaller than x = 5, the function values again become very large in size
since the denominator is near 0. Since the x-values are slightly less
than x = 5, the factor (x + 2) is a positive number, while the factor
(x – 5) is negative. Since your numerator is negative and your
denominator is negative (the product of a positive and a negative
number), the function values produced from these x-values are again
positive. Sketching this information produces the middle part of the
graph.
Moving to the left-hand piece of the domain, you may select an x-value
to give a starting point. Any x-value less than x = –2 can be selected.
For example, using x = –4 produces the point
. As you move
from x = –4 toward x = –2, the factor (x + 2) becomes close to 0,
making the denominator small and the value of the function large in
size. Since the x-values you are using now are less than x = –2, (x + 2)
is a negative number, as is (x – 5). Their product is positive, so your
function value is the quotient of the numerator, –4, divided by a
positive denominator. This quotient is a negative number that is
becoming large in size as you move toward x = –2.
89
90 CHAPTER 5: RATIONAL FUNCTIONS
As you move toward the other end of this part of the domain, you start
using x-values that are very far from 0, and negative. You determine
the values of the function by dividing –4 by the product of two negative
numbers that are very large in size, (x + 2) and (x – 5). The values you
obtain are negative and very close to 0 since the denominator is far
larger than the numerator. The graph of the function on the extreme
left-hand side of the domain is very close to, but below the x-axis. The
x-axis, y = 0, is an asymptote to the graph of f3(x) for these x-values.
Next you obtain the graph for the right-hand piece. You might again
select an x-value to use to obtain a starting point. For example, you
might use x = 7. This gives you the point
. If you move to the
left, toward x = 5, your denominator becomes small since (x – 5) is near
0, and the function values are large in size. The numerator is negative,
and since x is greater than 5 in this part of the domain, both (x + 2) and
(x – 5) are positive. The function values are negative, since you obtain
them by dividing a negative numerator by a positive denominator. The
graph for x-values slightly above x = 5 shows a function taking on very
low negative values near x = 5, rising to pass through
.
Finally, check what happens for large positive x-values. Both (x – 2)
and (x + 5) are very large positive numbers, so your function values are
the results of dividing –4 by large positive numbers. These values are
close to 0, and negative. The graph shows a curve approaching the
x-axis asymptotically, from below the axis. This completes the graph.
Some general principles:
When you divide a number by a denominator that is small in absolute
value, that is, near zero, your quotient is very large in size, although this
quotient could be either positive or negative.
When you divide a number by one much larger than it, the quotient is
small in absolute value. If the power of the denominator is greater than
that of the numerator, the quotient will be near zero when you use large
values for your variable.
To determine the value of a quotient when your x-values are very large
(in either positive or negative directions), you consider only the highestpower terms in both numerator and denominator.
CHAPTER 5: RATIONAL FUNCTIONS
Other examples of rational functions:
f4(x) =
Again there are two x-values that would make the denominator 0, both
x = –1 and x = 1. You must remove these two numbers from the
domain. The graph is in three pieces, from the extreme left to x = -1,
between x = –1 and x = 1, and from x = 1 to the extreme right.
You may select an x-value less than –1 to find a starting point for the
left-hand piece. Graph this point.
Extremely negative x-values give quotients of positive numbers divided
by positive numbers, so the function values on the left are positive.
Also, since for these inputs it is the highest degree terms that are
significant, you can ignore the effect of subtracting 1 in the
denominator.
=2
You can see that the function values are close to y = 2 on the extreme
left. Notice that your starting point has a function value higher than 2.
This is because the numerator is always more than twice as large as the
denominator. This shows you that your graph is above y = 2 while
close to this value for x-values far to the left.
As your x-values get closer to x = –1 (from below), the quotient (still
positive) gets larger and larger since the numerator is near 2, while the
denominator becomes small (near 0).
The graph for this section of the domain shows a curve above y = 2,
very close to y = 2 on the extreme left, passing through your starting
point, and becoming very large in the positive direction when your xvalues get near x = –1.
On the other side of x = –1, the denominator is still near 0, but now is
negative in sign. The numerator is still close to 2. Your quotient is
again very large in size, as always when you divide by numbers near 0,
but it is now negative.
The intercept is (0, 0), which gives you a point in the section between
x = –1 and x = 1.
91
92 CHAPTER 5: RATIONAL FUNCTIONS
Using values approaching x = 1 from below, you are again dividing a
number near 2 by a negative number close to 0. Such quotients are
again negative numbers of large size.
The graph in this middle section shows a curve that is very negative
near x = –1 that rises to pass through the intercept at the origin, and
again becoming extremely negative when you get near x = 1.
In the third section, input values just above x = 1 produce quotients that
are numbers near 2 divided by very small positive numbers. These
quotients are large positive numbers.
If you select an x-value greater than 1 to use for a starting point, you
can plot this now. Notice that the function value here is larger than 2.
When you use very large positive x-values you can again ignore the
small effect that the subtraction of 1 has in the denominator, and you
can again see that your function values get near 2 when you select large
enough x-values.
The graph of this third section begins with very high positive values
when you are just above x = 1, then passes through your point and
approaches y = 2 when your x-values get far enough to the right.
You might also note that this is the graph of an even function, so the
graph must be symmetric about the y-axis.
f5(x) =
.
The only x-value that makes the denominator of this function equal to
0 is the value x = 2. Exclude this value from your domain. You can
mark this exclusion with a vertical asymptote.
You can obtain the intercept as usual, the point
. When you
move from x = 0 to your left, toward more negative x-values, you see
that the denominator has larger sized numbers than the numerator. In
fact,
=
. Therefore, the end-behavior on the
extreme left is asymptotic to the x-axis. Since the x-values are negative
in this part of the domain, you can see that the curve is close to the xaxis but below the axis for extremely negative x-values.
CHAPTER 5: RATIONAL FUNCTIONS
When you use x-values very close to, but less than x = 2, the
denominator is near 0, but negative. The numerator, 3x2 – 3x – 36, is
also negative, near –30. The function values are therefore positive and
becoming larger as you take your x-values closer to x = 2.
The graph is near, but under, the x-axis for very negative x-values,
(
passes through the intercept 0, 4
1
2
) , then rises rapidly as the x-values
approach x = 2.
If the x-values you select are slightly above x = 2, then your numerator
is still near –30 and your denominator is close to 0, but positive, since
cubing an x-value larger than 2 produces a number larger than 8. The
result is negative, and large in size. When you use large values for x,
the higher power in the denominator causes the function values to be
near 0, but since the function is behaving like
, these values are
positive.
The graph of the function to the right of x = 2 shows extremely negative
values for x-values just above x = 2, then rises to cross the x-axis to
attain positive values, but the curve then heads toward the x-axis from
the positive side as the x-values increase.
f6(x) =
Since the denominator factors into 5(x2 – 2x – 3) = 5(x + 1)(x – 3), you
see that you must exclude x = –1 and x = 3 from the domain to prevent
getting a denominator of 0.
For x-values near x = –1, the function must get very large in size since
the denominator is near 0 for such x-values. If you use x-values less
than x = –1, both (x + 1) and (x – 3) are negative so the denominator is
positive for these x-values. Since the numerator isalways positive, the
function has very large positive values for x-values that are slightly
below x = –1.
For very large x-values,
=
. The function is
asymptotic to a parabola that opens upward and is fairly wide.
The graph on the left shows a curve that looks like a parabola opening
upward, then as you move across toward x = –1, the curve turns rapidly
upward.
93
94 CHAPTER 5: RATIONAL FUNCTIONS
For x-values slightly above x = –1, x + 1 is positive and x – 3 is
negative, so the denominator is negative and near 0. The function
(
values are large negative numbers. The intercept is 0, −
6
15
) . When
you use x-values slightly below x = 3, x + 1 is positive and x – 3 is
negative. The denominator for these x-values is negative and near 0,
so again the graph shows extremely negative values for x-values just
below x = 3. The graph between x = –1 and x = 3 rises from the
extremely negative values it has near x = –1, passes through the
intercept, and returns to very negative values when you approach x = 3.
For x-values greater than x = 3, the denominator is still near 0, but both
(x + 1) and (x – 3) are positive, so the function has large positive
values. As the x-values get larger, the function again approaches its
asymptotic curve, y =
. The graph shows high positive values when
the x-values are close to x = 3, decreases for a while, then heads toward
the parabola y =
.
When a function has a graph that is asymptotic to another curve, it is
important that you select your viewing window with care in order to have
a complete graph visible in your screen. If you view your graph from too
far away, the asymptotic curve will dominate the graph of the function and
you may lose the details of the graph that illustrate what is happening at
some of the x-values. In the graph for f 6(x) you have just completed, the
end behavior of the parabola
dominates the graph at the ends. If you
choose an x-interval for your graph that is too large, you will see the graph
of this parabola overwhelming the turns the graph makes between x = –1
and x = 3. The figure in the right-hand column shows what happens to the
graph when you use the x-interval [–100, 100].
Now that you understand how the graphs of rational functions behave, and
which features are the necessary to examine (particularly restrictions to the
domain and end-behavior), you can use your graphics calculator to produce
complete graphs of these functions. If you were graphing f 4(x), for
example, you would have to be sure that the left side boundary was lower
than x = –1, the right side above x = 1, the top higher than y = 2, and the
bottom should at least be low enough to show the curve crossing below the
axis. If your window was not at least this large, you might miss the graph
in one or more of the sections, and fail to obtain a complete graph.
X interval: [–100, 100]
CHAPTER 5: RATIONAL FUNCTIONS
Applications
The kinds of applied models that rational functions may describe usually
have domains that are naturally limited to certain sets of values.
For example, S(t) =
might model the starting salary for a
certain executive position (in thousands of dollars), where t is the
number of years that have passed since 1950. If you start to graph this
function, you will quickly realize that you must remove –75 from the
domain. It is not a possible value for t because it would produce a value
of 0 in the denominator. The model apparently should only be
considered for t-values above –75 (that is, after 1875).
The intercept gives the starting salary this position would have had in
1950 (0 years from 1950!), which is 7 ($7000).
As t increases (as time went by), the influence of the constants 1050 in
the numerator and 150 in the denominator diminish in importance in
calculation of the salary. By 1990 (t = 40), the starting salary had
become
, or $25,400. In the distant future, this salary will
be determined by
, which represents the ultimate
starting salary of $60,000.
Now the graph over this part of the domain shows a curve that is
extremely negative for t-values slightly above t = –75 (because for
values near here you are dividing a negative numerator by a small
positive denominator). A t-value of –75 makes sense, since it means a
time 75 years before 1950, or 1875, but the graph does not. Even in
1875, salaries were not negative! The model may make sense for some
values below 0, but probably not too far below. You would only use
this model for the part of the domain in which the results make sense,
maybe from t = –5 and higher.
When you are working with applied models, make sure you use the function
(and graph) only with values of the domain that give you realistic answers.
The mathematical model does not know which set of values this should be
and produces results (meaningful or not) throughout the entire domain of
the function. You must apply common sense to your results, which even
experienced workers sometimes forget to do.
95
96 CHAPTER 5: RATIONAL FUNCTIONS
Exercises
1. Sketch the complete graph of:
a. f(x) =
b. f(x) =
c. f(x) =
d. f(x) =
e. f(x) =
2. Sketch the complete graph of:
a. f(x) =
b. f(x) =
c. f(x) = –6 +
d. f(x) = 5 –
3. Write a possible equation for the function whose graph is given by:
a.
b.
c.
CHAPTER 5: RATIONAL FUNCTIONS
97
4. Sketch the complete graph of:
a. f(x) =
b. f(x) =
c. f(x) =
d. f(x) =
e. f(x) =
5. Sketch the complete graph (including all end behavior) of:
a. f(x) =
b. f(x) =
c. f(x) =
6. One model for the world's record in the long jump (in feet) is given by the function L(x) =
,
where x represents the number of years that have passed since 1900. Sketch a complete graph of this
model. What information does the intercept give you? The world's record in 1990 was 29.2 feet. What
does the model predict the record would be? According to this model, what is the ultimate world record?
7. The number of daily sales that a saleswoman can make is given by the function S(t) =
, where t
represents the number of weeks of sales experience she has. Sketch a complete graph for this function.
How many daily sales could she make when she first started the job? What will be the greatest number of
daily sales she can hope to achieve?
8. Find the maximum value of the function f (x) =
3
4
−2 + 8x − x
x
2
.
98 CHAPTER 5: RATIONAL FUNCTIONS
Answers
1.
a.
b.
d.
2.
a.
c.
e.
b.
d.
c.
CHAPTER 5: RATIONAL FUNCTIONS
3. Note that in each answer c represents some positive number.
a.
4.
b.
a.
c.
b.
Wide-range view of 4b.
c.
5.
a.
d.
Close-up view of 4b.
e.
b.
Wide-range view of 5b.
c.
Close-up view of 5b.
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100 CHAPTER 5: RATIONAL FUNCTIONS
6. The intercept notes that the world record in the long jump in 1900
was 24 feet. The model predicts a record of 29.83 feet, or 29 feet
10 inches, in 1990. The ultimate world record, according to the
model, is given by the asymptote, 30 feet.
7. With no experience, daily sales are 3 units, according to the model.
The number of sales, with great experience, will approach, but not
surpass, 20 units per day.
8.
The maximum point appears to be at (4.03, 15.87).