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In this section, you will:
  • Use arrow notation.
  • Solve applied problems involving rational functions.
  • Find the domains of rational functions.
  • Identify vertical asymptotes.
  • Identify horizontal asymptotes.
  • Graph rational functions.

Suppose we know that the cost of making a product is dependent on the number of items, x , produced. This is given by the equation C ( x ) = 15,000 x 0.1 x 2 + 1000. If we want to know the average cost for producing x items, we would divide the cost function by the number of items, x .

The average cost function, which yields the average cost per item for x items produced, is

f ( x ) = 15,000 x 0.1 x 2 + 1000 x

Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power.

In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator.

Using arrow notation

We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in [link] , and notice some of their features.

Graphs of f(x)=1/x and f(x)=1/x^2

Several things are apparent if we examine the graph of f ( x ) = 1 x .

  1. On the left branch of the graph, the curve approaches the x -axis ( y = 0 )   as   x .
  2. As the graph approaches x = 0 from the left, the curve drops, but as we approach zero from the right, the curve rises.
  3. Finally, on the right branch of the graph, the curves approaches the x- axis ( y = 0 )   as   x .

To summarize, we use arrow notation    to show that x or f ( x ) is approaching a particular value. See [link] .

Symbol Meaning
x a x approaches a from the left ( x < a but close to a )
x a + x approaches a from the right ( x > a but close to a )
x x approaches infinity ( x increases without bound)
x x approaches negative infinity ( x decreases without bound)
f ( x ) the output approaches infinity (the output increases without bound)
f ( x ) the output approaches negative infinity (the output decreases without bound)
f ( x ) a the output approaches a

Local behavior of f ( x ) = 1 x

Let’s begin by looking at the reciprocal function, f ( x ) = 1 x . We cannot divide by zero, which means the function is undefined at x = 0 ; so zero is not in the domain . As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in [link] .

x –0.1 –0.01 –0.001 –0.0001
f ( x ) = 1 x –10 –100 –1000 –10,000

We write in arrow notation

as  x 0 , f ( x )

As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in [link] .

x 0.1 0.01 0.001 0.0001
f ( x ) = 1 x 10 100 1000 10,000

We write in arrow notation

As  x 0 + ,   f ( x ) .

See [link] .

Graph of f(x)=1/x which denotes the end behavior. As x goes to negative infinity, f(x) goes to 0, and as x goes to 0^-, f(x) goes to negative infinity. As x goes to positive infinity, f(x) goes to 0, and as x goes to 0^+, f(x) goes to positive infinity.

This behavior creates a vertical asymptote , which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line x = 0 as the input becomes close to zero. See [link] .

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
is it a question of log
I rally confuse this number And equations too I need exactly help
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Commplementary angles
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
Practice Key Terms 5

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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