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Wize High School Grade 12 Pre-Calculus Textbook > Rational Functions

Graphing Rational Functions

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Graphing Rational Functions

Steps for graphing a rational function f(x)=axn+bxn−1+...cxm+dxm−1+...f(x)=\displaystyle\frac{ax^n+bx^{n-1}+...}{cx^m+dx^{m-1}+...}:
  1. Factor, if necessary
  2. Find the xx-intercepts & yy-intercepts
  3. Find the vertical asymptotes
  4. Find the horizontal asymptotes
  5. Find the missing points/holes
  6. Use a table of signs to determine if the function, f(x)f(x), will be above or below the horizontal asymptote. Any vertical asymptotes & x-intercepts will be used to determine the behaviour of f(x)f(x)
  7. Plot the function using the information found in Steps 1 - 6.

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Example

Graph f(x)=4x−5x−2f(x)=\displaystyle\frac{4x-5}{x-2}, stating the domain and range.

Step 1.
Factor, if necessary.

Factoring will not be required since f(x)=4x−5x−2f(x)=\displaystyle\frac{4x-5}{x-2} is in simplest form already.


Step 2.
Find the x-intercepts & y-intercepts.

The x-intercepts are determined by allowing f(x)=0f(x)=0:

f(x)=4x−5x−2(0=4x−5x−2)×(x−2)0=4x−5∴ x=54\begin{array}{rcl} f(x)&=&\displaystyle\frac{4x-5}{x-2}\\\\ \Bigg(0&=&\displaystyle\frac{4x-5}{x-2}\Bigg)\times(x-2)\\\\ 0&=&4x-5\\\\ \therefore{}~x&=&\displaystyle\frac{5}{4} \end{array}

The y-intercepts are determined by allowing x=0x=0:

f(x)=4x−5x−2f(x)=4(0)−50−2∴ f(x)=52\begin{array}{rcl} f(x)&=&\displaystyle\frac{4x-5}{x-2}\\\\ f(x)&=&\displaystyle\frac{4(0)-5}{0-2}\\\\ \therefore{}~f(x)&=&\displaystyle\frac{5}{2} \end{array}


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Step 3.
Find the vertical asymptotes.

The vertical asymptotes can found by finding the zeros of the denominator. In other words, set the denominator to 00 and solve.

Let x−2=0x-2=0.

The vertical asymptote is x=2.x=2.



Step 4.
Find the horizontal asymptotes.

The degree of the polynomial in the numerator (n=1) (n=1)~ is equivalent to the degree of the polynomial in the denominator (m=1)(m=1).


Therefore, the horizontal asymptote for f(x)f(x) is the ratio of the leading coefficients y=41=4y=\frac{4}{1}=4.


Step 5.
Find the missing points/holes.

There were no terms eliminated. Therefore, there are no missing points/holes.


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Step 6.
Use a table of signs to determine if the function, f(x)\green{f(x)}, will be above or below the horizontal asymptote. Any vertical asymptotes & x-intercepts will be used to determine the behaviour of f(x).\green{f(x).}


01.534x−5−++x−2−−+f(x)+−+\begin{array}{c|c|c|} &0&1.5&3\\\hline\\ 4x-5&-&+&+\\\\\hline\\ x-2&-&-&+\\\\\hline\\ f(x)&+&-&+\\\\ \end{array} Pick values of xx around the vertical asymptotes and x-intercpets

The function lies above the horizontal asymptote in the interval (−∞,54) & (2,∞)(-\infin, \frac{5}{4})~\&~(2,\infin)

The function lies below the horizontal asymptote in the interval (54,2)(\frac{5}{4},2)


Step 7.
Plot the function using the information found in Steps 1 - 6.


Domain: (−∞,2) ∪ (2,∞)(-\infin,2)~\cup~(2,\infin)

Range: (−∞,4) ∪ (4,∞)(-\infin,4)~\cup~(4,\infin)
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Example: Graphing Rational Functions

Sketch a graph of f(x)=xx2−5f(x)=\displaystyle\frac{x}{x^2-5}.

Step 1.
Factor, if necessary.

f(x)=xx2−5f(x)=x(x−5)(x+5)\begin{array}{rcl} f(x)&=&\displaystyle\frac{x}{x^2-5}\\\\ f(x)&=&\displaystyle\frac{x}{(x-\sqrt{5})(x+\sqrt{5})} \end{array}


Step 2.
Find the x-intercepts & y-intercepts.

The x-intercepts are determined by allowing f(x)=0f(x)=0:

f(x)=xx2−5(0=xx2−5)×(x2−5)0=x∴ x=0\begin{array}{rcl} f(x)&=&\displaystyle\frac{x}{x^2-5}\\\\ \Bigg(0&=&\displaystyle\frac{x}{x^2-5}\Bigg)\times(x^2-5)\\\\ 0&=&x\\\\ \therefore{}~x&=&0 \end{array}

The y-intercepts are determined by allowing x=0x=0:

f(x)=xx2−5f(x)=002−5∴ f(x)=0\begin{array}{rcl} f(x)&=&\displaystyle\frac{x}{x^2-5}\\\\ f(x)&=&\displaystyle\frac{0}{0^2-5}\\\\ \therefore{}~f(x)&=&0 \end{array}


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Step 3.
Find the vertical asymptotes.

The vertical asymptotes can found by finding the zeros of the denominator. In other words, set the denominator to 00 and solve.

Let (x−5)(x+5)=0(x-\sqrt{5})(x+\sqrt{5})=0.

The vertical asymptotes are x=±5x=\pm\sqrt{5}



Step 4.
Find the horizontal asymptotes.

The degree of the polynomial in the numerator (n=1) (n=1)~ is not equivalent to the degree of the polynomial in the denominator (m=2)(m=2).


Therefore, the horizontal asymptote for f(x)f(x) is the ratio of the leading coefficients y=0y=0.


Step 5.
Find the missing points/holes.

There were no terms eliminated. Therefore, there are no missing points/holes.


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Step 6.
Use a table of signs to determine if the function, f(x)\green{f(x)}, will be above or below the horizontal asymptote. Any vertical asymptotes & x-intercepts will be used to determine the behaviour of f(x).\green{f(x).}


−3−113x−−++x2−5+−−+f(x)−+−+\begin{array}{c|c|c|c|} &-3&-1&1&3\\\hline\\ x&-&-&+&+\\\\\hline\\ x^2-5&+&-&-&+\\\\\hline\\ f(x)&-&+&-&+\\\\ \end{array} Pick values of xx around the vertical asymptotes and x-intercpets

The function lies above the horizontal asymptote in the interval (−5,0) & (5,∞)(-\sqrt{5}, 0)~\&~(\sqrt{5},\infin)

The function lies below the horizontal asymptote in the interval (−∞,−5) & (0,5)(-\infin,-\sqrt{5})~\&~(0,\sqrt{5})


Step 7.
Plot the function using the information found in Steps 1 - 6.


Domain: (−∞,−5) ∪ (−5,5) ∪ (5,∞)(-\infin,-\sqrt{5})~\cup~(-\sqrt{5},\sqrt{5})~\cup~(\sqrt{5},\infin)

Range: (−∞,0) ∪ (0,∞)(-\infin,0)~\cup~(0,\infin)

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Example: Graphing Rational Functions

Sketch a graph of f(x)=x−3−2x−8f(x)=\displaystyle\frac{x-3}{-2x-8}.

Step 1.
Factor, if necessary.

f(x)=x−3−2x−8f(x)=x−3−2(x+4)\begin{array}{rcl} f(x)&=&\displaystyle\frac{x-3}{-2x-8}\\\\ f(x)&=&\displaystyle\frac{x-3}{-2(x+4)} \end{array}


Step 2.
Find the x-intercepts & y-intercepts.

The x-intercepts are determined by allowing f(x)=0f(x)=0:

f(x)=x−3−2(x+4)(0=x−3−2(x+4))×(−2(x+4))0=x−3∴ x=3\begin{array}{rcl} f(x)&=&\displaystyle\frac{x-3}{-2(x+4)}\\\\ \Bigg(0&=&\displaystyle\frac{x-3}{-2(x+4)}\Bigg)\times(-2(x+4))\\\\ 0&=&x-3\\\\ \therefore{}~x&=&3 \end{array}

The y-intercepts are determined by allowing x=0x=0:

f(x)=x−3−2x−8f(x)=0−3−2(0+4)f(x)=−3−2(4)∴ f(x)=38\begin{array}{rcl} f(x)&=&\displaystyle\frac{x-3}{-2x-8}\\\\ f(x)&=&\displaystyle\frac{0-3}{-2(0+4)}\\\\ f(x)&=&\displaystyle\frac{-3}{-2(4)}\\\\ \therefore{}~f(x)&=&\displaystyle\frac{3}{8} \end{array}


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Step 3.
Find the vertical asymptotes.

The vertical asymptotes can found by finding the zeros of the denominator. In other words, set the denominator to 00 and solve.

Let (x+4)=0(x+4)=0.

The vertical asymptote is x=−4x=-4



Step 4.
Find the horizontal asymptotes.

The degree of the polynomial in the numerator (n=1) (n=1)~ is equivalent to the degree of the polynomial in the denominator (m=1)(m=1).


Therefore, the horizontal asymptote for f(x)f(x) is the ratio of the leading coefficients y=−12y=\displaystyle-\frac{1}{2}.


Step 5.
Find the missing points/holes.

There were no terms eliminated. Therefore, there are no missing points/holes.

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Step 6.
Use a table of signs to determine if the function, f(x)\green{f(x)}, will be above or below the horizontal asymptote. Any vertical asymptotes & x-intercepts will be used to determine the behaviour of f(x).\green{f(x).}


−504x−3−−+−2−−−x+4−++f(x)−+−\begin{array}{c|c|c|} &-5&0&4\\\hline\\ x-3&-&-&+\\\\\hline\\ -2&-&-&-\\\\\hline\\ x+4&-&+&+\\\\\hline\\ f(x)&-&+&-\\\\ \end{array} Pick values of xx around the vertical asymptotes and x-intercpets

The function lies above the horizontal asymptote in the interval (−4,3)(-4,3)

The function lies below the horizontal asymptote in the interval (−∞,−4) & (3,∞)(-\infin,-4)~\&~(3,\infin)


Step 7.
Plot the function using the information found in Steps 1 - 6.


Domain: (−∞,−4) ∪ (−4,∞)(-\infin,-4)~\cup~(-4,\infin)

Range: (−∞,−12) ∪ (−12,∞)\Big(-\infin,\displaystyle-\frac{1}{2}\Big)~\cup~\Big(\displaystyle-\frac{1}{2},\infin\Big)
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Example: Graphing Rational Functions

Sketch a graph of f(x)=3x2+5x+22x−4f(x)=\displaystyle\frac{3x^2+5x+2}{2x-4}

Step 1.
Factor, if necessary.

f(x)=3x2+5x+22x−4f(x)=(3x+2)(x+1)2(x−2)\begin{array}{rcl} f(x)&=&\displaystyle\frac{3x^2+5x+2}{2x-4}\\\\ f(x)&=&\displaystyle\frac{(3x+2)(x+1)}{2(x-2)} \end{array}


Step 2.
Find the x-intercepts & y-intercepts.

The x-intercepts are determined by allowing f(x)=0f(x)=0:

f(x)=(3x+2)(x+1)2(x−2)(0=(3x+2)(x+1)2(x−2))×(2(x−2))0=(3x+2)(x+1)∴ x=−23, −1\begin{array}{rcl} f(x)&=&\displaystyle\frac{(3x+2)(x+1)}{2(x-2)}\\\\ \Bigg(0&=&\displaystyle\frac{(3x+2)(x+1)}{2(x-2)}\Bigg)\times(2(x-2))\\\\ 0&=&(3x+2)(x+1)\\\\ \therefore{}~x&=&-\displaystyle\frac{2}{3},~-1 \end{array}

The y-intercepts are determined by allowing x=0x=0:

f(x)=(3x+2)(x+1)2(x−2)f(x)=(3(0)+2)(0+1)2(0−2)f(x)=22(−2)∴ f(x)=−12\begin{array}{rcl} f(x)&=&\displaystyle\frac{(3x+2)(x+1)}{2(x-2)}\\\\ f(x)&=&\displaystyle\frac{(3(0)+2)(0+1)}{2(0-2)}\\\\ f(x)&=&\displaystyle\frac{2}{2(-2)}\\\\ \therefore{}~f(x)&=&\displaystyle-\frac{1}{2} \end{array}


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Step 3.
Find the vertical asymptotes.

The vertical asymptotes can found by finding the zeros of the denominator. In other words, set the denominator to 00 and solve.

Let (x−2)=0(x-2)=0.

The vertical asymptotes is x=2x=2



Step 4.
Find the horizontal (or oblique) asymptotes.

The degree of the polynomial in the numerator (n=2) (n=2)~ is not equivalent to the degree of the polynomial in the denominator (m=1)(m=1).


Therefore, there is an oblique asymptote.

Divide f(x)=3x2+5x+22x−4f(x)=\displaystyle\frac{3x^2+5x+2}{2x-4}

3522↓62231124\begin{array}{r|ccc} &3&5&2\\ 2&\downarrow&6&22\\\hline &3&11&24 \end{array}

Divide the quotient by 2:

(3x+11)÷2=3x2+112(3x+11)\div2=\displaystyle\frac{3x}{2}+\displaystyle\frac{11}{2}

The oblique asymptote is the quotient. So, y=3x2+112y=\displaystyle\frac{3x}{2}+\displaystyle\frac{11}{2}


Step 5.
Find the missing points/holes.

There were no terms eliminated. Therefore, there are no missing points/holes.


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Step 6.
Use a table of signs to determine if the function, f(x)\green{f(x)}, will be above or below the oblique asymptote. Any vertical asymptotes & x-intercepts will be used to determine the behaviour of f(x).\green{f(x).}


−2−0.75033x+2−−++x+1−−++2++++x−2−−−+f(x)−−−+\begin{array}{c|c|c|c|c} &-2&-0.75&0&3\\\hline\\ 3x+2&-&-&+&+\\\\\hline\\ x+1&-&-&+&+\\\\\hline\\ 2&+&+&+&+\\\\\hline\\ x-2&-&-&-&+\\\\\hline\\ f(x)&-&-&-&+\\\\ \end{array} Pick values of xx around the vertical asymptotes and x-intercepts

The function lies above the oblique asymptote in the interval (2,∞)(2,\infin)

The function lies below the oblique asymptote in the interval (−∞,2)(-\infin,2)


Step 7.
Plot the function using the information found in Steps 1 - 6.


Domain: (−∞,2) ∪ (2,∞)(-\infin,2)~\cup~(2,\infin)

Practice: Graphing Rational Functions

A rational function has the following properties:
  • A horizontal asymptote at y=0y=0.
  • Vertical asymptotes at x=−5,3x=-5, 3.
  • An x-intercept at x=0.8x=0.8
Which rational function best describes the above properties?

Practice: Graphing Rational Functions

Sketch a graph of f(x)=x+4−2x−6f(x)=\displaystyle\frac{x+4}{-2x-6}, stating all asymptotes, x-intercepts, missing points, and positive & negative intervals.

Practice: Graphing Rational Functions

Find a rational function of the form f(x)=ax+bcx+df(x)=\displaystyle\frac{ax+b}{cx+d} that describes the following:

  • Vertical asymptote at x=5x=5
  • Horizontal asymptote at y=3y=3
  • X-intercept at x=−25x=-\displaystyle\frac{2}{5}

Practice: Graphing Rational Functions

Write a rational function with the following properties:
  • An oblique asymptote at y=x+4y=x+4
  • A vertical asymptote at x=5x=5
  • One of the x-intercepts is at x=2x=2
Extra Practice
Rational Functions
Let f(x)=x2+5x+6x−4f(x)=\dfrac{x^2+5x+6}{x-4} be a rational function.
Graphing Rational Functions
Practice: Graphing Rational Functions
Write a rational function with the following properties:
An oblique asymptote at y=−3x+1y=-3x+1
Graphing Rational Functions
Practice: Graphing Rational Functions
Write a rational function with the following properties:
An oblique asymptote at y=2x−7y=2x-7
Graphing Rational Functions
Practice: Graphing Rational Functions
Find a rational function of the form f(x)=ax+bcx+df(x)=\displaystyle\frac{ax+b}{cx+d} that describes the following:
Vertical asymptote at x=−1x=-1
Graphing Rational Functions
Practice: Graphing Rational Functions
Find a rational function of the form f(x)=ax+bcx+df(x)=\displaystyle\frac{ax+b}{cx+d} that describes the following:
Vertical asymptote at x=−2x=-2
Graphing Rational Functions
Practice: Graphing Rational Functions
Sketch a graph of f(x)=x2−25x+5f(x)=\dfrac{x^2-25}{x+5}, finding all asymptotes, x-intercepts, missing points, and positive & negative intervals.
Graphing Rational Functions
Practice: Graphing Rational Functions
Sketch a graph of f(x)=−x+9x+3f(x)=\dfrac{-x+9}{x+3}, finding all asymptotes, intercepts, missing points, and positive & negative intervals.
Rational Functions
Practice: Rational Functions
Match the rational function with the correct sketch of its graph.