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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


Example 1


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


Image illustrating Rational Functions




  • 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.




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:


Image illustrating Example


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.



Your answers

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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}

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Extra Practice