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Wize AP Calculus (BC) Textbook > Review: Functions/Pre-Calculus

Rational Functions

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Rational Functions

A rational function can be expressed as:
f(x)=axn+bxn−1+...cxm+dxm−1+...\boxed{f(x)=\displaystyle\frac{ax^n+bx^{n-1}+...}{cx^m+dx^{m-1}+...}}
where nn is the largest exponent in the numerator and mm is the largest exponent in the denomiator.

Properties of Rational Functions

  • X-Intercepts can be found by setting f(x)=0f(x)=0 and solving for xx
  • Vertical asymptotes can be found by setting the denominator to 0 and solving for xx
  • Horizontal asymptotes:
  • If n<m‾\underline{n<m}: There is a horizontal asymptote at y=0y=0
  • If n=m‾\underline{n=m}: There is a horizontal asymptote at y=acy=\displaystyle\frac{a}{c}
  • If n>m‾\underline{n>m}: There is no horizontal asymptote. There may be an oblique/diagonal asymptote.
  • If factoring the numerator & denominator of f(x)f(x), any terms that cancel out indicates where a missing point/hole in f(x)f(x)
  • The domain is the set of all real numbers except where there are vertical asymptotes & missing points/holes
  • The range is the set of all real numbers except where there are horizontal asymptotes in most cases
Watch Out!
Rational functions may only attain values strictly above or below horizontal asymptotes. This will affect the range.

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Example 1

Let's look at f(x)=2x−3x+1f(x)=\displaystyle\frac{2x-3}{x+1}.




  • xx-Intercept: (32,0)\Bigg(\displaystyle\frac{3}{2},0\Bigg)
  • Vertical Asymptote: x=−1x=-1
  • Horizontal Asymptote: y=2y=2
  • Missing Points/Holes: None
  • Intervals of Increasing: (−∞,−1) ∪ (−1,∞)(-\infin,-1)~\cup~(-1,\infin)
  • Intervals of Decreasing: NA
  • Domain: (−∞,−1) ∪ (−1,∞)(-\infin,-1)~\cup~(-1,\infin)
  • Range: (−∞,2) ∪ (2,∞)(-\infin,2)~\cup~(2,\infin)
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Solving Rational Equations

Rational equations are a type of equation containing at least one rational term, f(x)g(x)\boxed{\displaystyle\frac{f(x)}{g(x)}}, where f(x) & g(x)f(x)~\&~g(x) are continuous polynomials.


Rational equations can contain non-permissible values (NPV) that identify the restrictions of the domain. They identify the value of the variable that gives a 0 0~in the denominator.


Wize Tip
The NPV's are equivalent to the vertical asymptote and missing points/holes.

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Example

Solve 2x−1=5x+2\displaystyle\frac{2}{x-1}=\displaystyle\frac{5}{x+2}, stating any NPV's.


Step 1.
Identify any non-permissible values.

The NPV's are the vertical asymptotes.
Therefore, x≠−2,1x\neq{-2,1} otherwise there is a total of 00 in the denominator.


Step 2.
Find the lowest common denominator and multiply each term by it.

The lowest common denominator is (x−1)(x+2)(x-1)(x+2).

(2x−1=5x+2)×(x−1)(x+2)   →    2(x+2)=5(x−1)\Bigg(\displaystyle\frac{2}{x-1}=\displaystyle\frac{5}{x+2}\Bigg)\times(x-1)(x+2)~~~{\color{red}\rightarrow}~~~~2(x+2)=5(x-1)


Step 3.
Solve for ′x′.\green{'x'.}

2(x+2)=5(x−1)2x+4=5x−53x=9∴ x=3\begin{array}{} 2(x+2)&=&5(x-1)\\\\ 2x+4&=&5x-5\\\\ 3x&=&9\\\\ \therefore~x&=&3 \end{array}


Step 4.
Verify.

2x−1=5x+223−1=53+222=551=1✓\begin{array}{} \displaystyle\frac{2}{x-1}&=&\displaystyle\frac{5}{x+2}\\\\ \displaystyle\frac{2}{3-1}&=&\displaystyle\frac{5}{3+2}\\\\ \displaystyle\frac{2}{2}&=&\displaystyle\frac{5}{5}\\\\ 1&=&1&\checkmark \end{array}
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Example: Solving Rational Equations

Solve 3xx+5+1x−2=7x2+3x−10\displaystyle\frac{3x}{x+5}+\frac{1}{x-2}=\displaystyle\frac{7}{x^2+3x-10}, stating any NPV's.


Step 1.
Identify any non-permissible values.

The NPV's are the vertical asymptotes.

(x+5)=0→∴ x=−5(x−2)=0→∴ x=2x2+3x−10=(x+5)(x−2)=0→∴ x=−5,2\begin{array}{rccccccc} (x+5)&=&0&&\rightarrow&&\therefore~x&=&-5\\\\ (x-2)&=&0&&\rightarrow&&\therefore~x&=&2\\\\ x^2+3x-10=(x+5)(x-2)&=&0&&\rightarrow&&\therefore~x&=&-5, 2 \end{array}


Therefore, x≠−5,2x\neq{-5,2} otherwise there is a total of 00 in the denominator.


Step 2.
Find the lowest common denominator and multiply each term by it.

The lowest common denominator is (x+5)(x−2)(x+5)(x-2).

(3xx+5+1x−2=7x2+3x−10)×(x+5)(x−2)      →      3x(x−2)+(x+5)=7\Bigg(\displaystyle\frac{3x}{x+5}+\frac{1}{x-2}=\displaystyle\frac{7}{x^2+3x-10}\Bigg)\times(x+5)(x-2)~~~~~~{\color{red}\rightarrow}~~~~~~3x(x-2)+(x+5)=7


Step 3.
Solve for ′x′.\green{'x'.}

3x(x−2)+(x+5)=73x2−6x+x+5=73x2−5x−2=0(3x2−6x)+(x−2)=0(3x+1)(x−2)=0∴ x=−13,2\begin{array}{rcl} 3x(x-2)+(x+5)&=&7\\\\ 3x^2-6x+x+5&=&7\\\\ 3x^2-5x-2&=&0\\\\ (3x^2-6x)+(x-2)&=&0\\\\ (3x+1)(x-2)&=&0 \\\\\therefore~x&=&-\displaystyle\frac{1}{3}, 2 \end{array}

According to the NPV's, x≠2x\neq2.

Therefore, x=2x=2 is an extraneous solution.


Step 4.
Verify.

3xx+5+1x−2=7x2+3x−103(−13)(−13)+5+1(−13)−2=7(−13)2+3(−13)−10−314−37=−914−914=−914         ✓\begin{array}{rcl} \displaystyle\frac{3x}{x+5}+\frac{1}{x-2}&=&\displaystyle\frac{7}{x^2+3x-10}\\\\ \displaystyle\frac{3(-\frac{1}{3})}{(-\frac{1}{3})+5}+\frac{1}{(-\frac{1}{3})-2}&=&\displaystyle\frac{7}{(-\frac{1}{3})^2+3(-\frac{1}{3})-10} \\\\-\displaystyle\frac{3}{14}-\displaystyle\frac{3}{7}&=&-\displaystyle\frac{9}{14}\\\\ -\displaystyle\frac{9}{14}&=&-\displaystyle\frac{9}{14}~~~~~~~~~\checkmark \end{array}


Therefore, x=−13x=-\displaystyle\frac{1}{3} is the only solution.

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Example: Rational Functions

A graph of the function f(x)=x2−6x+8x2−3x−4f(x)=\displaystyle{\frac{x^2-6x+8}{x^2-3x-4}} is sketched below:


Identify the equations for any vertical & horizontal asymptotes, and missing points/holes.

Factor f(x)f(x):

f(x)=x2−6x+8x2−3x−4f(x)=(x−4)(x−2)(x−4)(x+1)f(x)=x−2x+1\begin{array}{rcl} f(x)&=&\displaystyle{\frac{x^2-6x+8}{x^2-3x-4}}\\\\ f(x)&=&\displaystyle{\frac{(x-4)(x-2)}{(x-4)(x+1)}}\\\\ f(x)&=&\displaystyle{\frac{x-2}{x+1}} \end{array}


Vertical Asymptotes:

Let x+1=0x+1=0.

The vertical asymptote is at x=−1.x=-1.


Horizontal Asymptote:

The degree of the numerator is equivalent to the degree in the denominator.

Therefore, y=1y=1.


Missing Points/Holes

Missing points/holes exist whenever terms in the numerator and denominator cancel.

Since the term (x−4)(x-4) was eliminated, there is a missing point/hole at x=4.x=4.

Given the equation 2x−3−34−x=2x−2x2−x−12\displaystyle\frac{2}{x-3}-\frac{3}{4-x}=\displaystyle\frac{2x-2}{x^2-x-12}, answer the following questions.



Solve for xx.


Identify the vertical asymptotes, the horizontal asymptotes, the x-intercepts, and any missing points of:

f(x)=x2−11x+18x2−3x+2f(x)=\displaystyle\frac{x^2-11x+18}{x^2-3x+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
A rational function has the following properties:
A horizontal asymptote at y=0y=0.
Graphing Rational Functions
Practice: Graphing Rational Functions
A rational function has the following properties:
A horizontal asymptote at y=0y=0.
Rational Functions
Practice: Rational Functions
Match the equation for the rational function with its appropriate properties.
Rational Functions
Practice: Rational Functions
Match the equation for the rational function with its appropriate properties.
Rational Functions
Practice: Rational Functions
Identify the vertical asymptotes, the horizontal asymptotes, the x-intercepts, and any missing points of:
f(x)=3x2+14x+8x2−2x−14f(x)=\dfrac{3x^2+14x+8}{x^2-2x-14}
Rational Functions
Practice: Rational Functions
Identify the vertical asymptotes, the horizontal asymptotes, the x-intercepts, and any missing points of:
f(x)=2x2+x−12x2+5x−3f(x)=\dfrac{2x^2+x-1}{2x^2+5x-3}
Rational Functions
Practice: Rational Functions
Match the rational function with the correct sketch of its graph.
Rational Functions
Practice: Rational Functions
Match the rational function with the correct sketch of its graph.