This section covers:
 Revisiting Direct and Inverse Variation
 Polynomial Long Division
 Asymptotes of Rationals
 Transformations of Rational Functions
 Drawing Rational Graphs — General Rules
 Finding Rational Functions from Graphs, Points, Tables, or Sign Charts
 Applications of Rational Functions
 More Practice
Again, Rational Functions are just those with polynomials in the numerator and denominator, so they are the ratio of two polynomials. Now that we know how to work with both rationals and polynomials, we’ll work on more advanced solving and graphing with them.
Note that Rational Inequalities, including Absolute Values, can be found here. Also, since limits exist with Rational Functions and their asymptotes; limits are discussed here in the Limits and Continuity section.
Revisiting Direct and Inverse Variation
We went over Direct, Inverse, Joint and Combined Variation here, but little did we know that we were working with Rational Functions when we were solving Inverse or Combined variation problem! This is because we had variables in the denominators for these types of problems.
Polynomial Long Division
We need to take a minute (sorry!) and talk about long division with polynomials. Long division with polynomials is sometimes needed when the degree (highest exponent in any variable) in the numerator is larger than the degree of the denominator. If the denominator is just one term (a monomial like \(8x\)), we just put each term in the numerator over the denominator. This is also called simplifying or reducing the fraction:
Math 
Notes 
Regular Fractions: \(\displaystyle \frac{{3+71}}{8}\,=\,\frac{3}{8}+\frac{7}{8}+\frac{1}{8}\,=\,\frac{3}{8}+\frac{7}{8}\frac{1}{8}\)
Polynomial Fractions: \(\require {cancel} \displaystyle \begin{align}\frac{{8{{y}^{5}}3{{y}^{4}}2y8}}{{{{y}^{3}}}}&=\frac{{8{{y}^{5}}}}{{{{y}^{3}}}}\frac{{3{{y}^{4}}}}{{{{y}^{3}}}}\frac{{2y}}{{{{y}^{3}}}}\frac{8}{{{{y}^{3}}}}\\&=\frac{{8{{y}^{{{{{\cancel{5}}}^{2}}}}}}}{{{{y}^{{\cancel{3}}}}}}\frac{{3{{y}^{{{{{\cancel{4}}}^{1}}}}}}}{{{{y}^{{\cancel{3}}}}}}\frac{{2y}}{{{{y}^{3}}}}\frac{8}{{{{y}^{3}}}}\\&=8{{y}^{2}}3y+\frac{{2y8}}{{{{y}^{3}}}}\end{align}\) 
Just like with regular fractions, if we want to break up the fraction, we can put each term of the numerator over the denominator. In the case of the polynomial, we can subtract the exponents when we divide; if the degree (exponent) of the top is less than the degree of the bottom, we have to leave it as a fraction.
As you can see, any terms with exponents that are smaller on the top are put together and can be thought of as a “remainder”. In this example, the \(\displaystyle \frac{{2y8}}{{{{y}^{3}}}}\) is the remainder of \(\displaystyle \frac{{8{{y}^{5}}3{{y}^{4}}2y8}}{{{{y}^{3}}}}\). Watch signs! 
When there are more than two terms on the bottom, it gets a little more complicated, and we have to do polynomial long division. There’s actually an easier way to do this with Synthetic Division, which we’ll learn about later, but let’s work with long division first. It’s really cool, since we can divide polynomials very similar to “regular” numbers. Notice how the steps line up:
Regular Math Division  Polynomial Long Division  




Let’s do more polynomial long division. Notice if we are missing a term in the dividend part (under the division sign), we have to create one with a coefficient of 0, just so we can line up things when we do the dividing.
Math  Notes 



Asymptotes of Rationals
Because rational functions typically have variables in the denominator, graphing them can be a bit tricky. We’ll introduce here the notion of an asymptote, or a graph that gets closer and closer to a line but never hits it. (It comes from a Greek word, meaning “not falling together”.) We will learn later that asymptotes are examples of limits; meaning that something gets closer and closer to a number, without actually touching it.
The reason we see asymptotes in rationals is because, again, there are typically \(x\) values (domains) where the function or graph does not exist at all, since we can’t divide by “0”.
One of the simplest rational functions, the inverse function (as seen in the Parent Functions and Transformations section), is \(\displaystyle y=\frac{1}{x}\):
Notice how, as \(x\) gets larger and larger, \(y\) gets closer and closer to 0. This is because as \(x\) gets larger and larger, we’ll get a smaller and smaller fraction (convince yourself by putting \(\displaystyle \frac{1}{{10000000}}\) in your calculator), but \(x\) can never be 0, since we can’t have a 0 in the denominator. This is a type of a Infinite Limit.
In this graph, we have horizontal asymptotes at “\(y=0\)” and vertical asymptotes at “\(x=0\)”. Horizontal asymptotes are also called end behavior asymptotes, since they occur when \(x\) gets very small and also very big. This is a type of a Limit at Infinity. I like to call \((0,0)\) the “anchor point” of the graph, since it’s the point where the two asymptotes “intersect”. Points \(\left( {1,1} \right)\) and \(\left( {1,1} \right)\) are called “reference points” of the graph, since they are at the “corners” of the branches.
Vertical asymptotes are sometimes written as VA, and end behavior asymptotes are written as EBA. Again, end behavior asymptotes are called such since they exist at the extreme areas of the \(x\): where \(x=\infty \) or \(x=\infty \). Horizontal asymptotes (also written as HA) are a special type of end behavior asymptotes.
Transformations of Rational Functions
Again, the parent function for a rational (inverse) function is \(\displaystyle y=\frac{1}{x}\), with horizontal and vertical asymptotes at \(x=0\) and \(y=0\), respectively. As in other functions, we can perform vertical or horizontal stretches, flips, and/or left or right shifts. Here are more inverse function graphs that you may have to draw and also shift or transform:
Negative Inverse Function  Inverse Squared Function  Negative Inverse Squared Function 
In the transformed function, \(\displaystyle y=a\frac{1}{{\frac{1}{b}\left( {xh} \right)}}+k\), the inverse function is vertically stretched by \(a\) units, horizontally stretched by \(b\) units, shifted right \(h\) units, and up \(k\) units. Note that a horizontal stretch is the same as the reciprocal of a vertical stretch!
The new “anchor point” would be at \(\left( {h,k} \right)\), and we would have horizontal and vertical asymptotes at \(x=h\) and \(y=k\), respectively. The new “reference points” (“corners” of the graph branches) would be at \(\left( {b+h,a+k} \right)\) and \(\left( {b+h,a+k} \right)\) instead of at \(\left( {1,1} \right)\) and \(\left( {1,1} \right)\), respectively. If the rational function were negative, there would be a flip around the \(x\)axis (before any vertical shifts!), and, if there are no horizontal or vertical shifts, this would be the same as flipping around the \(y\)axis, since the function is odd.
Again, we always perform the stretches and flips first, with the horizontal and vertical shifts to follow. (We can also perform these transformations using tcharts, like we learned in the Parent Functions and Transformations section).
Let’s do some examples. The graph on the right shows what happen when we shift the graph of \(\displaystyle y=\frac{1}{x}\) “2 units to the right, and 3 units up”. The “anchor point” of the graph to the right is \((2,3)\), and the asymptotes are \(x=2\) (VA) and \(y=3\) (EBA/HA).
Parent Inverse Function 
Shifted Inverse Function 
Here are a few more examples:
Transformed Inverse Function 
Transformed Inverse Squared Function 
\(\displaystyle \frac{1}{x}\) function, vertical stretch of 3 (points will start at \(\left( {1,3} \right),\left( {1,3} \right)\) instead of \(\left( {1,1} \right),\left( {1,1} \right)\)), flipped around the \(x\)axis, shifted to the left 1, shifted down 4:  \(\displaystyle \frac{1}{{{{x}^{2}}}}\) function, vertical stretch of 4 (points will start at \(\left( {1,4} \right),\left( {1,4} \right)\) instead of \(\left( {1,1} \right),\left( {1,1} \right)\)), shifted to the right 2, shifted down 6: 
Note that by using long division, we can change a rational function in a form such as \(\displaystyle g\left( x \right)=\frac{{ax+b}}{{cx+d}}\) to a rational function that can be transformed using the techniques above.
Here is an example:
Rational Function  Transformation and Graph  
Graph \(\displaystyle g(x)=\frac{{6x+2}}{{x1}}\) by using long division and transformations of the rational parent function \(f(x)=\displaystyle \frac{1}{x}\).
First use long division: \(\displaystyle \begin{array}{c}x1\overset{{6+\frac{8}{{x1}}}}{\overline{\left){\begin{array}{l}\,6x+2\,\,\,\,\\\,\underline{{6x6}}\\\,\,\,\,\,\,\,\,\,\,\,\,8\end{array}}\right.}}\,\\\,\\g\left( x \right)=8\left( {\frac{1}{{x1}}} \right)+6\end{array}\)

Graph, using the tchart method, that we learned in the Parent Functions and Transformations (“normal operations for the \(y\), opposite operations for the \(x\), using PEMDAS):
We could also see that the “center point” of the function is at \(\left( {1,6} \right)\), and there’s a vertical stretch of 8 (go over 1 and up/down 8 from that point). 
Continuous Versus Discontinuous Functions
We talked about continuous versus discontinuous (or noncontinuous) functions in the Piecewise Functions Section here, but notice that the functions above are discontinuous, meaning that you have to lift up your pencil when you draw them from left to right.
Continuous Functions are just that; they could theoretically be drawn from “left to right” (or less commonly “down to up”) without picking up a pencil. Any linear function, for example, is continuous. Some rational functions are also continuous, as we’ll see later.
Drawing Rational Graphs – General Rules
We can look at more complicated forms of rational functions and, from just a small set of rules, roughly draw the graph of that function – it’s like magic ;)! We may need a tchart to help us out, but we’ll be able to graph most rational functions pretty quickly.
The table below shows rules and examples. You’ll find that these same rules apply to the graphs above (after finding a common denominator and combining terms if necessary), but usually you are taught those graphs separately.
Remember that the degree of the polynomial is the highest exponent of any of the terms, and if the polynomial is in factored form, you have to multiply the variables (or add the exponents and find the highest exponent) to find the degree. For example, the degree of \({{x}^{3}}+2x+4\) is 3 because of the \({{x}^{3}}\), and the degree of \(({{x}^{2}}+3)({{x}^{3}}2)\) is 5, since \({{x}^{2}}\times {{x}^{3}}={{x}^{5}}\).
Rules for Graphing Rationals 
Example 
First factor both the numerator and denominator and cross out any factors in both the numerator and denominator.
⇒ If any of these factors contain variables, these are removable discontinuities, or “holes” and will be little circles on the graphs. The idea is that if you cross out a polynomial, you can’t forget that it was in the denominator, and any value(s) making the denominator 0 must be eliminated. (We will see this in a graph later.) To get the \(y\) value for the discontinuity, plug in the \(x\) value after you cross out the factors.
⇒ The domain of a rational function is all real numbers, except those that make the denominator equal 0, as we saw earlier.
(Note that if after you cross out factors and you still have that same factor that you crossed out left on the bottom, the “hole” will turn into a vertical asymptote; follow the rules below). 
\(\require{cancel} \displaystyle y=\frac{{2{{x}^{2}}+5x3}}{{x\left( {{{x}^{2}}9} \right)}}=\frac{{\cancel{{\left( {x+3} \right)}}\left( {2x1} \right)}}{{x\cancel{{\left( {x+3} \right)}}\left( {x3} \right)}}=\frac{{2x1}}{{x\left( {x3} \right)}}\)
After factoring, we can cross out \((x+3)\) on the top and bottom, so there is a removable discontinuity (a little circle on the graph) where \(x=3\) and \(\displaystyle y=\frac{{2\left( {3} \right)1}}{{\left( {3} \right)\left( {33} \right)}}=\frac{{7}}{{18}}=\frac{7}{{18}}\) (plug –3 in for \(y\) in reduced fraction). The hole is at \(\displaystyle \left( {3,\frac{7}{{18}}} \right)\).
Domain is \(\left( {\infty ,3} \right)\cup \left( {3,0} \right)\cup \left( {0,3} \right)\cup \left( {3,\infty } \right)\), since we have to “skip over” values that could make the denominator 0. 
To get vertical asymptotes or VAs:
⇒ After determining if there are any holes in the graph, factor (if necessary) what’s left in the denominator and set the factors to 0. For any value of \(x\) where these factors could be 0, this creates a vertical asymptote at “\(x=\)” for these values.
⇒ Don’t forget to include the factors with “\(x\)” alone (\(x=0\) is the vertical asymptote). Note: There could a multiple number of vertical asymptotes, or no vertical asymptotes. If there are no vertical asymptotes, or holes, the rational function is continuous. 
\(\displaystyle y=\frac{{2{{x}^{2}}+5x3}}{{x\left( {{{x}^{2}}9} \right)}}=\frac{{\cancel{{\left( {x+3} \right)}}\left( {2x1} \right)}}{{x\cancel{{\left( {x+3} \right)}}\left( {x3} \right)}}=\frac{{2x1}}{{x\left( {x3} \right)}}\)
After we’ve gotten rid of the hole, we look at what’s left on the bottom; these are vertical asymptote(s). We see that vertical asymptotes occur when \(\left( {x0} \right)=0\,\,\,\text{or}\,\,\,\left( {x3} \right)=0\), or \(\displaystyle x=0\,\,\,\text{or}\,\,\,x=3\).
Again, domain is \(\left( {\infty ,3} \right)\cup \left( {3,0} \right)\cup \left( {0,3} \right)\cup \left( {3,\infty } \right)\), since anything that could make the denominator 0 (even a hole) can’t be included. We have to “skip over” –3, 0, and 3. 
To get the end behavior asymptote (EBA), compare the degree in the numerator to the degree in the denominator. There can be at most 1 EBA and, most of the time, these are horizontal.
⇒ If the degree (largest exponent) on the bottom is greater than the degree on the top, the EBA (which is also a horizontal asymptote or HA) is \(y=0\). 
\(\displaystyle y=\frac{{x+2}}{{{{x}^{2}}4}}\)
Notice that even though we can take out a removable discontinuity (\(x+2\)), the bottom still has a higher degree than the top, so the HA/EBA is \(y=0\). 
⇒ If the degree on the top is greater than the degree on the bottom, there is no HA. However, if the degree on the top is one more than the degree on the bottom, than there is a slant (oblique) EBA asymptote, which is discussed below. 
\(\displaystyle y=\frac{{{{x}^{3}}+2}}{{x4}}\)
No HA/EBA. Vertical asymptote is still \(\displaystyle x=4\). 
⇒ If the degree is the same on the top and the bottom, divide coefficients of the variables with the highest degree on the top and bottom; this is the HA/EBA. You can determine this asymptote even without factoring. 
\(\displaystyle y=\frac{{2{{x}^{3}}+2}}{{3{{x}^{3}}4}}\)
Since the degree on the top and bottom are both 3, the HA/EBA is \(\displaystyle y=\frac{2}{3}\). 
⇒ If the degree on the top is one more than the degree on the bottom, then the function has a slant or oblique EBA in the form \(y=mx+b\). We have to use long division to find this linear equation.
We can just ignore or “throw away” the remainder and just use the linear equation. Weird, huh? 
\(\displaystyle y=\frac{{2{{x}^{2}}+x+1}}{{x4}}\) \(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2x\,\,+9\\x4\overline{\left){{2{{x}^{2}}+x+1}}\right.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{2{{x}^{2}}8x\,\,\,\,\,\,}}\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9x+1\end{array}\) EBA: \(y=2x+9\) 
⇒ (more advanced) Find the point where any horizontal asymptotes cross the function by setting the function to the horizontal asymptote, and solving for “\(x\)”. You already have the “\(y\)” (from the HA equation).  Q: Where does \(\displaystyle y=\frac{{{{x}^{2}}+x}}{{{{x}^{2}}+x12}}\) intersect its EBA?
A: Note that the EBA/HA is \(\displaystyle y=\frac{{1}}{1}=1\). Now set \(\displaystyle \frac{{{{x}^{2}}+x}}{{{{x}^{2}}+x12}}=1\) and cross multiply: \({{x}^{2}}+x=1\left( {{{x}^{2}}+x12} \right);\,\,\,\,2x=12;\,\,\,x=6\) . The point where the function intersects the EBA/HA is \((6,1)\). 
The way I like to remember the horizontal asymptotes (HAs) is: BOBO BOTN EATS DC (Bigger On Bottom, asymptote is 0; Bigger On Top, No asymptote; Exponents Are The Same, Divide Coefficients).
Note that there can be multiple vertical asymptotes, but only one EBA (HA or slant/oblique) asymptote. Note that also the function can intersect the EBA asymptote, but not intercept the vertical asymptote(s). Also, sometimes the function intersects the EBA and then come back up or down to get closer to the asymptote.
To create rational graphs without a calculator:
 Factor to see if any removable discontinuities (or holes) exist; cross out on top and bottom. To get the \(y\)value of the hole, use the crossedout version and plug in the \(x\)value. You know that part of the curve of the graph goes close to that point, but you have to graph a small circle there.
 Draw any vertical asymptotes (VAs) (nonremovable discontinuities) from setting anything left in the denominator to 0. (You may get none, and there can be more than one.) Note that if there are no removable discontinuities or vertical asymptotes, the function is continuous. Any VAs divide the graph into sections; you’ll be drawing a graph in each section.
 Draw any end behavior asymptotes (EBA: HA or oblique) from BOBO BOTN EATS DC and oblique/slant asymptote instructions above. (You may get none, but there will be at most one). With slant (oblique) asymptotes, the curves will slant, but still hug the asymptote.
 Determine the \(x\)intercepts (where \(y=0\)), and \(y\)intercepts (where \(x=0\)).
 (More advanced) See if the function crosses any horizontal asymptotes by setting the original function equal to the HA. Solve for \(x\); you already have the \(y\) (from the asymptote).
 Draw “tcharts” to fill in extra “key” points, for example, on the sides of the EBA asymptotes.
 Domain is everything except where the removable discontinuities or asymptotes exist.
 Have your graphs “hug every asymptote”, but remember that you will never have more than one point on a vertical line, since we’re drawing functions.
 Also note that if any VA’s (denominator) are repeated, they are said to have a multiplicity. For VA’s with an odd multiplicity (odd exponent, including no exponent), the arrows will alternate on either side of the VA (opposite corners); this looks more like the \(\displaystyle y=\frac{1}{x}\) graph. For even multiplicity (even exponent), the arrows will go in the same direction on either side of the VA (side by side); this looks more like the \(\displaystyle y=\frac{1}{{{{x}^{2}}}}\) graph. But check points when graphing!
 If you have two or more VA‘s, the graph can look either like a “parabola” or a “cubic” inside the two asymptotes with outside curves on either the top or bottom. (They are not parabolas or cubics; they just look them. Always check points, using tcharts, if necessary!)
Remember that, for horizontal asymptotes, \(\underset{{x\to \infty }}{\mathop{{\lim }}}\,f\left( x \right)=\text{EBA}\), and \(\underset{{x\to \infty }}{\mathop{{\lim }}}\,f\left( x \right)=\text{EBA}\). This means that as \(x\) gets closer and closer to positive and negative infinity, \(x\) gets closer and closer to the horizontal asymptote. And always check points to make sure you’ve graphed correctly!
Here’s a quick problem to check the understanding of graphing rational functions without using numbers!
Graphing Rationals Problem  Solution 
If \(\displaystyle y=\frac{{a\left( {x+b} \right)\left( {xc} \right)}}{{d\left( {xc} \right)\left( {x+e} \right)}}\), what are the following, in terms of \(\displaystyle a,b,c,d,\,\,\text{and}\,e\)?
a) \(x\)intercept(s) b) \(y\)intercept c) Vertical asymptote(s) d) End behavior asymptote e) Domain f) \(x\)value of any removable discontinuities g) Is the function continuous or discontinuous? 
a) The \(x\)intercept is the value when \(y=0\); after factoring out the hole (removable discontinuity) at \(x=c\); we have at \(x+b=0;\,\,\,x=b:\,\,\left( {b,0} \right)\).
b) The \(y\)intercept is the value when \(x=0\); it must be \(\displaystyle \frac{{a\left( {0+b} \right)\cancel{{\left( {0c} \right)}}}}{{d\cancel{{\left( {0c} \right)}}\left( {0+e} \right)}}=\frac{{ab}}{{de}}:\,\left( {0,\frac{{ab}}{{de}}} \right)\). c) The vertical asymptote (VA) is found after factoring out the hole and setting the denominator to 0; we have \(x+e=0:\,\,x=e\). d) The end behavior asymptote (EBA) is \(\displaystyle y=\frac{a}{d}\), since the degree is the same on the top and bottom (divide coefficients). e) The domain is all real numbers except \(x\ne c,\,\,x\ne e\), since these values would make the denominator 0. f) The \(x\)value of the removable discontinuity (hole) is \(x=c\). g) The function is discontinuous, since it has a hole (and vertical asymptote). 
Here are some “real” examples of Graphing Rational Functions:
Graphing Rational Function Steps  Graph  
\(\displaystyle y=\frac{{3{{x}^{2}}8x16}}{{3{{x}^{2}}16x+16}}\)
Factor: \(\require{cancel} \displaystyle y=\frac{{\left( {3x+4} \right)\cancel{{\left( {x4} \right)}}}}{{\left( {3x4} \right)\cancel{{\left( {x4} \right)}}}}\)
Removable Discontinuity or Hole: \(x=4\), plug in 4 for \(x\) to get \(y=2\). RD is \((4,2)\). VA: Set denominator to 0 after removing the hole: \(\displaystyle 3x4=0;\,\,\,\,\,x=\frac{4}{3}\) 
EBA/HA: Since the degree is the same on the top and bottom (both are 2), we take the coefficients and divide them: \(\displaystyle y=\frac{3}{3}=1\).
\(x\)intercept (root): Set \(y\) (or top) to 0 after “crossing out” hole: \(\displaystyle 3x+4=0;\,\,\,\,x=\frac{4}{3}\,;\,\,\,\left( {\frac{4}{3},0} \right)\) \(y\)intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{3(0)+4}}{{3(0)4}}=1\,\,\,\,\,\left( {0,1} \right)\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \(\displaystyle \left( {\infty ,\frac{4}{3}} \right)\cup \left( {\frac{4}{3},4} \right)\cup \left( {4,\infty } \right)\) 
\(\displaystyle y=\frac{x}{{{{x}^{2}}+4x5}}\)
Factor: \(\displaystyle y=\frac{x}{{\left( {x+5} \right)\left( {x1} \right)}}\)
Removable Discontinuity or Hole: none VA: Set denominator to 0 after factoring; we have 2 of them: \(x=5,\,\,\,\,\,x=1\) 
EBA/HA: Since the degree on the top (1) is less than the degree on the bottom (2), the EBA or HA is \(y=0\).
\(x\)intercept (root): Set \(y\) (or top) to 0: \(x=0:\,\,\,\,\left( {0,0} \right)\) \(y\)intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{\left( 0 \right)}}{{{{{\left( 0 \right)}}^{2}}+4\left( 0 \right)5}}=0:\,\,\left( {0,0} \right)\) “tchart”: Try some points around the vertical asymptotes: \(\begin{array}{l}x=6,\,\,y\approx .86\\x=4,\,\,y=.8\\x=2,\,\,y\approx .29\end{array}\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \(\left( {\infty ,5} \right)\cup \left( {5,1} \right)\cup \left( {1,\infty } \right)\) 
\(\displaystyle y=\frac{{{{x}^{2}}}}{{{{x}^{2}}+4x5}}\) (similar to above, but with \({{x}^{2}}\) on top)
Factor: \(\displaystyle y=\frac{{{{x}^{2}}}}{{\left( {x+5} \right)\left( {x1} \right)}}\)
Removable Discontinuity or Hole: none VA: Set denominator to 0 after factoring; we have 2 of them: \(x=5,\,\,\,\,\,x=1\) 
EBA/HA: Since the degree is the same on the top and bottom (both are 1), we take the coefficients and divide them: \(\displaystyle y=\frac{1}{1}=1\).
\(x\)intercept (root): Set \(y\) (or top) to 0: \({{x}^{2}}=0:\,\,\left( {0,0} \right)\) \(y\)intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{{{{\left( 0 \right)}}^{2}}}}{{{{{\left( 0 \right)}}^{2}}+4\left( 0 \right)5}}=0:\,\,\,\left( {0,0} \right)\) “tchart”: Try some points around the vertical asymptotes: \(\begin{array}{l}x=6,\,\,y\approx 5.14\\x=4,\,\,y=3.2\\x=2,\,\,y\approx .57\end{array}\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \(\left( {\infty ,5} \right)\cup \left( {5,1} \right)\cup \left( {1,\infty } \right)\) 
Here’s one with an even multiplicity in the denominator; notice how the graphs hug the vertical asymptote from the same direction:
Graphing Rational Function Steps 
Graph 

\(\displaystyle y=\frac{{3{{x}^{2}}12x}}{{{{x}^{2}}+2x+1}}\)
Factor: \(\displaystyle y=\frac{{3x\left( {x4} \right)}}{{{{{\left( {x+1} \right)}}^{2}}}}\)
Removable Discontinuity or Hole: none VA: Set denominator to 0: \(x+1=0;\,\,\,\,x=1\) 
EBA/HA: Since the degree is the same on the top and bottom (both are 2), we take the coefficients and divide them: \(\displaystyle y=\frac{3}{1}=3\).
\(x\)intercept (roots): Set \(y\) to 0: \(\begin{array}{l}x=0:\,\,\left( {0,0} \right)\\x=4:\,\,\left( {4,0} \right)\end{array}\) \(y\)intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{3\left( 0 \right)\left( {04} \right)}}{{{{{\left( {0+1} \right)}}^{2}}}}=0:\,\,\left( {0,0} \right)\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \((\infty ,1)\cup (1,\infty )\).
See how the graph hugs the vertical asymptote \(x=1\) from the same direction (positive \(\infty \)), since the multiplicity of the exponent is 2 (even)? 
Here are more advanced examples, with a slant (oblique) asymptote, a passthrough asymptote, and one with what looks like a “hole” (but actually a vertical asymptote):
Graphing Rational Function Steps  Graph  
\(\displaystyle y=\frac{{3{{x}^{2}}4x7}}{{x4}}\)
Factor: \(\displaystyle y=\frac{{\left( {3x7} \right)\left( {x+1} \right)}}{{\left( {x4} \right)}}\) EBA: Note that the degree on the top (2) is one more than the degree on the bottom (1), so the EBA is a slant or oblique asymptote. Use long division to see that the EBA is \(y=3x+8\): \(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x\,\,\,\,+8\\x4\overline{\left){{3{{x}^{2}}4x7}}\right.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{3{{x}^{2}}12x\,\,\,\,\,\,}}\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8x7\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{8x32\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{N/A}\end{array}\) 
VA: Set denominator to 0: \(x4=0:\,\,\,\,x=4\)
\(x\)intercepts (roots): Set \(y\) (or top) to 0: \(\displaystyle \begin{align}3x7=0;\,\,\,x=\frac{7}{3}:\,\,\,\,\,\,\left( {\frac{7}{3},0} \right)\\x+1=0;\,\,\,\,x=1:\,\,\,\,\,\left( {1,0} \right)\end{align}\) \(y\)intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{3{{{\left( 0 \right)}}^{2}}4\left( 0 \right)7}}{{\left( 0 \right)4}}=\frac{7}{4}:\) \(\displaystyle \left( {0,\,\,\frac{7}{4}} \right)\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \(\left( {\infty ,4} \right)\cup \left( {4,\infty } \right)\) 
\(\displaystyle y=\frac{{x4}}{{{{x}^{3}}+8}}\)
VA: Set denominator to 0: \({{x}^{3}}+8=0;\,\,\,{{x}^{3}}=8;\,\,\,x=2\)
EBA/HA: Since the degree on the bottom is greater than the degree on the top, the EBA/HA is \(y=0\).

\(x\)intercept (root): Set \(y\) (or top) to 0:
\(x4=0;\,\,x=4:\,\,\left( {4,0} \right)\) \(y\)Intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{04}}{{{{{(0)}}^{3}}+8}}=\frac{1}{2}:\,\left( {0,\,\,\frac{1}{2}} \right)\)
Note that the HA (\(y=0\)) crosses the function at: \(\displaystyle \frac{{x4}}{{{{x}^{3}}+8}}=0;\,\,\,\,x=4:\,\left( {4,0} \right)\)
“tchart”: Try some points around the vertical asymptotes: \(\begin{array}{c}x=3,\,\,\,y=.37\\x=1,\,\,\,y=.71\end{array}\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \(\displaystyle \left( {\infty ,2} \right)\cup \left( {2,\infty } \right)\) 
\(\displaystyle y=\frac{{{{x}^{2}}2x}}{{{{x}^{2}}4x+4}}\)
Factor: \(\require{cancel} \displaystyle y=\frac{{x\cancel{{\left( {x2} \right)}}}}{{\left( {x2} \right)\cancel{{\left( {x2} \right)}}}}\)
Removable Discontinuity or Hole: \(x=2\), plug in 2 for \(x\) to get … OOPS! Doesn’t work, so there’s no removable discontinuity, but a vertical asymptote at \(x=2\). VA: Only vertical asymptote is at \(x=2\). 
EBA/HA: Since the degree is the same on the top and bottom (both are 1), we take the coefficients and divide them: \(\displaystyle y=\frac{1}{1}=1\).
\(x\)intercept (root): After crossing out factors, set \(y\) to 0: \(x=0:\,\,\,\left( {0,0} \right)\) \(y\)Intercept: Set \(x\) to 0: \(\displaystyle y=\frac{{{{0}^{2}}2\left( 0 \right)}}{{{{0}^{2}}4\left( 0 \right)+4}}=0:\,\,\left( {0,0} \right)\) 
Domain: Can’t be any value of \(x\) that makes the bottom zero: \((\infty ,2)\cup (2,\infty )\) 
Note that if you end up with a line after taking out the removable discontinuity, the EBA is actually considered that line (could be a trick test question): 🙄
Graphing Rational Function Steps  Graph  
\(\displaystyle y=\frac{{2{{x}^{2}}2x}}{{x1}}\)
Factor: \(\require{cancel} \displaystyle y=\frac{{2x\cancel{{\left( {x1} \right)}}}}{{\cancel{{x1}}}}=2x\)
Removable Discontinuity or Hole: \(x=1,\) plug in 1 for \(x\) to get \(y=2\). RD is \((1,2)\). EBA: The degree on the top (2) is one more than the degree on the bottom (1), so the EBA is a slant or oblique asymptote. We don’t need to use long division, since we can see that the slant EBA is \(y=2x\). 
VA: none
\(x\)intercept (root): Set \(y\) to 0: \(0=2\left( x \right);\,\,\,\,x=0:\,\,\,\left( {0,0} \right)\) \(y\)intercept: Set \(x\) to 0. Do this after crossing out hole: \(y=2\left( 0 \right)=0:\,\,\left( {0,0} \right)\)

Domain: Can’t be any value of \(x\) that makes the bottom zero: \((\infty ,1)\cup (1,\infty )\) 
Finding Rational Functions from Graphs, Points, Tables, or Sign Charts
You may be asked to look at a rational function graph and find a possible equation from a rational function graph or a table of points:
Rational Function: Find Equation from Graph  
Determine Equation:
Removable Discontinuity or Hole: We see that there is a removable discontinuity at \(\left( {3,2} \right)\), so we could have \(\left( {x3} \right)\) in both the numerator and denominator. EBA: We see that the EBA is an HA at \(y=4\). We know that the degree of the numerator is the same as the denominator, and we would get –4 if we divided the coefficients. We’ll see this below when we check for the \(y\)intercept. VA: We see that the VA is \(x=1\). We could have an \(\left( {x1} \right)\) in the denominator. \(x\)intercept: We see that the \(x\)intercept is \(\left( {2,0} \right)\); we know that 2 is a root (the \(x\) that makes \(y=0\)). We could have an \(\left( {x2} \right)\) in the numerator. \(y\)intercept: We see that the \(y\)intercept is \(\left( {0,8} \right)\); we know that \(x\) is 0 when \(y\) is –8. So far we have: \(\displaystyle y=\frac{{\left( {x3} \right)\left( {x2} \right)}}{{\left( {x3} \right)\left( {x1} \right)}}\), and the degrees of the top and bottom look good. If we plug in 0 for \(x\), we get \(\displaystyle y=\frac{{\left( {03} \right)\left( {02} \right)}}{{\left( {03} \right)\left( {01} \right)}}=\frac{{\left( {3} \right)\left( {2} \right)}}{{\left( {3} \right)\left( {1} \right)}}=2\). But \(y\) should equal –8, so we have to multiply the function by –4 (which lines up with the EBA!). Therefore, a possible equation for the graph could be: \(\displaystyle y=\frac{{4\left( {x3} \right)\left( {x2} \right)}}{{\left( {x3} \right)\left( {x1} \right)}}\) Try it; it works! \(\displaystyle \surd \) 
Problem:
Here’s a problem where we need to write a possible rational equation from a table:
\(x\)  –100  –50  –1.1  –1  –.9  0  2.9  3  3.1  50  100 
(approximate) \(y\)  3.03  3.06  33  Error  –27  0  2.23  Error  2.27  2.94  2.97 
Solution:
It looks like there’s a hole (removable discontinuity) at \(x=3\), since the \(y\)values at either end are close and the same sign. It also looks like there’s a vertical asymptote at \(x=1\), since the \(y\)values at either end have opposite signs. So far, we have \(\displaystyle f\left( x \right)=\frac{{\left( {x3} \right)}}{{\left( {x3} \right)\left( {x+1} \right)}}\), but since the function goes through \(\left( {0,0} \right)\), the function must look more like \(\displaystyle f\left( x \right)=\frac{{x\left( {x3} \right)}}{{\left( {x3} \right)\left( {x+1} \right)}}\).
Try \(\displaystyle f\left( {100} \right):\,\,f\left( {100} \right)=\frac{{100\left( {1003} \right)}}{{\left( {1003} \right)\left( {100+1} \right)}}\approx .99\). But the table shows the point \(\left( {100,2.97} \right)\), with a \(y\) 3 times what we got! Thus, \(\displaystyle f\left( x \right)=\frac{{3x\left( {x3} \right)}}{{\left( {x3} \right)\left( {x+1} \right)}}\) would work. Tricky!
Problem:
Here’s one more where we have to write a rational inequality based on a sign chart (sign pattern).
Write a rational inequality for the sign chart below:
Solution:
We see that the critical points are –7, –5, –1, 0, and 3. Thus, we should have factors \(\left( {x+7} \right),\,\left( {x+5} \right),\,\left( {x+1} \right),\,x\) and \(\left( {x3} \right)\) in our rational function.
The critical point where there’s a sign change and an open hole (–5) must be a vertical asymptote; the critical points where’s there’s an open hole and no sign change (3) must be a removable discontinuity (hole). (To see this, graph some rational functions on your calculator). The rest of the factors (roots) must be in the numerator, and there must be a “bounce” at \(x=0\) (no sign change). Thus, we can build the rational expression \(\displaystyle \frac{{{{x}^{2}}\left( {x+7} \right)\left( {x+1} \right)\left( {x3} \right)}}{{\left( {x+5} \right)\left( {x3} \right)}}\).
Now we need to know if this rational inequality represented by this sign chart is \(\le 0\) or \(\ge 0\); we need to include the “or equal to” since the factors on top contain closed circles.
At this point, I would just plug in a number, such as –2 into the expression and see if it’s positive or negative; it’s negative: \(\displaystyle \frac{{{{{\left( {2} \right)}}^{2}}\left( {2+7} \right)\left( {2+1} \right)\left( {23} \right)}}{{\left( {2+5} \right)\left( {23} \right)}}=\frac{{20}}{3}\le 0\). The sign chart shows this interval to be positive, so the expression \(\displaystyle \frac{{{{x}^{2}}\left( {x+7} \right)\left( {x+1} \right)\left( {x3} \right)}}{{\left( {x+5} \right)\left( {x3} \right)}}\) works! (To make this into an inequality, we’d have \(\displaystyle \frac{{{{x}^{2}}\left( {x+7} \right)\left( {x+1} \right)\left( {x3} \right)}}{{\left( {x+5} \right)\left( {x3} \right)}}\ge 0\), which is the same as \(\displaystyle \frac{{{{x}^{2}}\left( {x+7} \right)\left( {x+1} \right)\left( {x3} \right)}}{{\left( {x+5} \right)\left( {x3} \right)}}\le 0\). Try it with other numbers in the other intervals; it works!
Applications of Rational Functions
We did some of application earlier here in the Rational Functions, Equations and Inequalities section, but here’s another that deals more with the concept of asymptotes:
Problem:
The concentration of a drug is monitored in the bloodstream of a patient. The drug’s concentration \(C\left( t \right)\) can be modeled by \(\displaystyle C\left( t \right)=\frac{{5t}}{{{{t}^{2}}+1}}\), where \(t\) is in hours, and \(C\left( t \right)\) is in mg.
a) What is the equation of the horizontal asymptote associated with this function? What does it mean about the drug’s concentration in the patient’s bloodstream as time increases?
b) When does the maximum concentration of the drug occur, and what is this maximum concentration?
Solution:
a) The asymptote of the function is \(y=0\), since the degree on the bottom is greater than the degree on the top. What this means is that, as time goes on, the drug is basically negligible in the patient; its concentration gets closer and closer to 0 mg.
b) To find the maximum concentration, put the equation in the graphing calculator and use the maximum function to find both the \(x\) and \(y\) values. You can see that the maximum concentration of 2.5 mg occurs after 1 hour:
Instructions  Screen 
Push “Y=” and enter the equation 5x/(x^2+1) in “Y_{1}=“. (You may need to press CLEAR after “Y_{1}=” first to clear out older equations). For the \({{t}^{2}}\) part, either use the “x^{2}” button or use “x^2“. Unless you have a very old version of the calculator Operating System, after the “x^2“, you have to use the right cursor button to move the cursor back down after putting in the exponent.
Push GRAPH. You may need to hit “ZOOM 6” (ZoomStandard) and “ZOOM 0” (ZoomFit) or maybe scroll up or down to “ZOOM A” (ZQuadrant1 – so you won’t have negative numbers) to make sure you see the maximum in the graph.
You can also use the WINDOW button to change the minimum and maximum values of your \(x\) and \(y\) values; you’ll just want to play around with the values until you can see the maximum in the first quadrant. 

To get the maximum, push “2^{nd} TRACE” (CALC), and then either push 3 for MINIMUM or 4 for MAXIMUM. In our case, we’ll push 4, or move the cursor down to MAXIMUM.
The calculator will say “Left Bound?” and then, using the cursors, move the cursor anywhere to the left of the maximum point and hit ENTER. Ignore the \(x\) and \(y\) values. When the calculator says “Right Bound?” move the cursor anywhere to the right of the maximum point and hit ENTER. Ignore the \(x\) and \(y\) values. The calculator will say “Guess?”.
Hit ENTER once more, and you have your maximum point, which is \((1,2.5)\). 
Learn these rules and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Graphing and Finding Roots of Polynomial Functions — you are ready!