2.13 Rational Functions – Partial FractionsHL ONLY

1. Characteristics of Rational Functions

A rational function is a function of the form $$f(x)=\dfrac{p(x)}{q(x)},\qquad q(x)\neq 0,$$ where $p(x)$ and $q(x)$ are polynomials. The domain is obtained by excluding all values of $x$ that make the denominator zero.

Rational functions are studied mainly through their undefined values, intercepts, and asymptotic behavior.

  1. Domain: solve $q(x)=0$, then exclude those values from $\mathbb{R}$.
  2. Vertical Asymptotes (VA): after cancelling any common factors, vertical asymptotes usually occur at the remaining zeros of the denominator.
  3. Holes: if a factor cancels between numerator and denominator, then the function may have a removable discontinuity, called a hole, rather than a vertical asymptote.
  4. $x$-intercepts: solve $p(x)=0$, but only after making sure the denominator is not zero.
  5. $y$-intercept: evaluate $f(0)$, provided $0$ belongs to the domain.
  6. Horizontal Asymptotes (HA): compare the degrees of $p(x)$ and $q(x)$:
    • If $\deg p(x)\lt \deg q(x)$, then the horizontal asymptote is $y=0$.
    • If $\deg p(x)=\deg q(x)$, then $$y=\dfrac{\text{leading coefficient of }p(x)}{\text{leading coefficient of }q(x)}.$$
    • If $\deg p(x)\gt \deg q(x)$, then there is no horizontal asymptote.

NOTICE

  • A rational function is undefined when its denominator is zero.
  • Do not automatically call every denominator zero a vertical asymptote; check first whether a factor cancels.
  • A graph may cross its horizontal asymptote at some finite value of $x$.
  • If the numerator degree is exactly one greater than the denominator degree, an oblique asymptote may occur.

EXAMPLE 1 (Determining Domain, Intercepts, and Asymptotes)

Analyze $$f(x)=\dfrac{7x^2+5x-3}{x^2-3x+2}.$$

Solution:

Factor the denominator: $$\begin{aligned} x^2-3x+2&=(x-1)(x-2). \end{aligned}$$ Hence the domain is $$x\in\mathbb{R},\qquad x\neq 1,\quad x\neq 2.$$
Since no factor cancels with the numerator, the vertical asymptotes are $$x=1\qquad\text{and}\qquad x=2.$$
The degrees of the numerator and denominator are equal. Therefore the horizontal asymptote is the ratio of the leading coefficients: $$y=\dfrac{7}{1}=7.$$
To find the $y$-intercept, evaluate $f(0)$: $$f(0)=\dfrac{-3}{2}.$$ Therefore the $y$-intercept is $$(0,-\dfrac{3}{2}).$$
To find the $x$-intercepts, solve the numerator equation: $$\begin{aligned} 7x^2+5x-3&=0,\\ x&=\dfrac{-5\pm\sqrt{5^2-4(7)(-3)}}{2(7)},\\ x&=\dfrac{-5\pm\sqrt{109}}{14}. \end{aligned}$$ Therefore the $x$-intercepts are $$\left(\dfrac{-5+\sqrt{109}}{14},0\right) \qquad\text{and}\qquad \left(\dfrac{-5-\sqrt{109}}{14},0\right).$$

2. Range of Rational Functions

To find the range of a rational function algebraically, set $y=f(x)$ and rearrange the equation into a polynomial equation in $x$. Then determine for which values of $y$ real solutions for $x$ exist.

For many rational functions involving quadratics, this leads to a quadratic equation in $x$. A real value of $x$ exists precisely when the discriminant is non-negative: $$\Delta\geq 0.$$

EXAMPLE 2 (A Rational Function with Range $\mathbb{R}$)

Analyze the range of $$f(x)=\dfrac{x^2-6x+8}{x^2-4x+3}.$$

Solution:

Factor the numerator and denominator: $$\begin{aligned} x^2-6x+8&=(x-2)(x-4),\\ x^2-4x+3&=(x-1)(x-3). \end{aligned}$$ Therefore, $$f(x)=\dfrac{(x-2)(x-4)}{(x-1)(x-3)}.$$
The denominator is zero at $$x=1\qquad\text{and}\qquad x=3,$$ so the vertical asymptotes are $$x=1\qquad\text{and}\qquad x=3.$$
Since the numerator and denominator have the same degree, the horizontal asymptote is $$y=1.$$
The $x$-intercepts are given by the zeros of the numerator: $$x=2\qquad\text{and}\qquad x=4.$$ Hence the $x$-intercepts are $$(2,0)\qquad\text{and}\qquad (4,0).$$
The $y$-intercept is $$f(0)=\dfrac{8}{3},$$ so the $y$-intercept is $$(0,\dfrac{8}{3}).$$
To find the range, let $$y=\dfrac{x^2-6x+8}{x^2-4x+3}.$$ Multiply through by the denominator: $$\begin{aligned} y(x^2-4x+3)&=x^2-6x+8,\\ yx^2-4yx+3y&=x^2-6x+8,\\ (y-1)x^2+(-4y+6)x+(3y-8)&=0. \end{aligned}$$
This is a quadratic equation in $x$. For real solutions to exist, the discriminant must satisfy $$\Delta\geq 0.$$ Compute the discriminant: $$\begin{aligned} \Delta&=(-4y+6)^2-4(y-1)(3y-8),\\ &=16y^2-48y+36-4(3y^2-11y+8),\\ &=16y^2-48y+36-12y^2+44y-32,\\ &=4y^2-4y+4,\\ &=4(y^2-y+1). \end{aligned}$$
Since $$y^2-y+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\gt 0,$$ it follows that $$\Delta=4(y^2-y+1)\gt 0$$ for all real values of $y$.
Therefore every real value of $y$ is attained, so the range is $$y\in\mathbb{R}.$$
Graph of $f(x)=\dfrac{x^2-6x+8}{x^2-4x+3}$
Vertical asymptotes: $x=1,\;x=3$
Horizontal asymptote: $y=1$
$x$-intercepts: $(2,0),\;(4,0)$
$y$-intercept: $(0,\dfrac{8}{3})$

EXAMPLE 3 (A Rational Function with a Restricted Range)

Analyze the intercepts and range of $$f(x)=\dfrac{x^2-3x-4}{x^2-4x+3}.$$

Solution:

Factor the numerator and denominator: $$\begin{aligned} x^2-3x-4&=(x-4)(x+1),\\ x^2-4x+3&=(x-1)(x-3). \end{aligned}$$ Therefore, $$f(x)=\dfrac{(x-4)(x+1)}{(x-1)(x-3)}.$$
There are no common factors, so the vertical asymptotes are $$x=1\qquad\text{and}\qquad x=3.$$ Since the degrees are equal, the horizontal asymptote is $$y=1.$$
The $x$-intercepts are found from the zeros of the numerator: $$x=-1\qquad\text{and}\qquad x=4.$$ Thus the $x$-intercepts are $$(-1,0)\qquad\text{and}\qquad (4,0).$$
The $y$-intercept is $$f(0)=\dfrac{-4}{3},$$ so the $y$-intercept is $$(0,-\dfrac{4}{3}).$$
To find the range, set $y=f(x)$: $$y=\dfrac{x^2-3x-4}{x^2-4x+3}.$$ Rearranging gives $$\begin{aligned} y(x^2-4x+3)&=x^2-3x-4,\\ yx^2-4yx+3y&=x^2-3x-4,\\ (y-1)x^2+(-4y+3)x+(3y+4)&=0. \end{aligned}$$
For real solutions in $x$, the discriminant must satisfy $\Delta\geq 0$: $$\begin{aligned} \Delta&=(-4y+3)^2-4(y-1)(3y+4),\\ &=16y^2-24y+9-4(3y^2+y-4),\\ &=4y^2-28y+25. \end{aligned}$$
Therefore, $$4y^2-28y+25\geq 0.$$ The boundary values are $$\begin{aligned} y&=\dfrac{28\pm\sqrt{(-28)^2-4(4)(25)}}{2(4)},\\ &=\dfrac{28\pm\sqrt{384}}{8},\\ &=\dfrac{28\pm 8\sqrt{6}}{8},\\ &=\dfrac{7}{2}\pm\sqrt{6}. \end{aligned}$$
Hence the range is $$y\in\left(-\infty,\dfrac{7}{2}-\sqrt{6}\right] \cup \left[\dfrac{7}{2}+\sqrt{6},\infty\right).$$ Numerically, $$y\in(-\infty,1.05]\cup[5.95,\infty).$$
Graph of $f(x)=\dfrac{x^2-3x-4}{x^2-4x+3}$
Vertical asymptotes: $x=1,\;x=3$
Horizontal asymptote: $y=1$
$x$-intercepts: $(-1,0),\;(4,0)$
$y$-intercept: $(0,-\dfrac{4}{3})$

3. Oblique Asymptotes

If the numerator degree is exactly one greater than the denominator degree, that is $$\deg p(x)=\deg q(x)+1,$$ then the rational function may have an oblique asymptote.

Polynomial long division writes the function as $$f(x)=Ax+B+\dfrac{r(x)}{q(x)},$$ where $r(x)$ is the remainder. Since $$\dfrac{r(x)}{q(x)}\to 0\qquad\text{as}\qquad x\to\pm\infty,$$ the graph approaches the line $$y=Ax+B.$$

EXAMPLE 4 (Finding an Oblique Asymptote)

Determine the complete asymptotic structure of $$f(x)=\dfrac{4x^2-12x+1}{2x-6}.$$

Solution:

The denominator is zero when $$2x-6=0\quad\Rightarrow\quad x=3.$$ Therefore the vertical asymptote is $$x=3.$$
The numerator has degree $2$ and the denominator has degree $1$. Since the numerator degree is greater than the denominator degree, there is no horizontal asymptote.
Use polynomial division: $$\begin{aligned} \dfrac{4x^2-12x+1}{2x-6} &=2x+\dfrac{1}{2x-6}. \end{aligned}$$
Since $$\dfrac{1}{2x-6}\to 0\qquad\text{as}\qquad x\to\pm\infty,$$ the oblique asymptote is $$y=2x.$$
Graph of $f(x)=\dfrac{4x^2-12x+1}{2x-6}$
Vertical asymptote: $x=3$
Horizontal asymptote: none
Oblique asymptote: $y=2x$
$y$-intercept: $(0,-\dfrac{1}{6})$

4. Partial Fractions

Partial fractions rewrite a rational expression as a sum of simpler rational expressions. This is useful in integration, algebraic manipulation, and curve analysis.

Partial fractions are applied to proper rational expressions, where $$\deg p(x)\lt\deg q(x).$$ If the rational expression is improper, first use polynomial division.

  • Distinct linear factors: $$\dfrac{p(x)}{(x-r_1)(x-r_2)} = \dfrac{A}{x-r_1} + \dfrac{B}{x-r_2}.$$
  • Repeated linear factors: $$\dfrac{p(x)}{(x-r)^2} = \dfrac{A}{x-r} + \dfrac{B}{(x-r)^2}.$$
  • Irreducible quadratic factors: $$\dfrac{p(x)}{(x-r)(x^2+px+q)} = \dfrac{A}{x-r} + \dfrac{Bx+C}{x^2+px+q}.$$

EXAMPLE 5 (Distinct Linear Factors)

Decompose $$\dfrac{3x-5}{x^2-4x+3}$$ into partial fractions.

Solution:

Factor the denominator: $$x^2-4x+3=(x-1)(x-3).$$
Set $$\dfrac{3x-5}{(x-1)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x-3}.$$
Multiply both sides by $(x-1)(x-3)$: $$3x-5=A(x-3)+B(x-1).$$
Use strategic substitution. Let $x=3$: $$\begin{aligned} 3(3)-5&=B(3-1),\\ 4&=2B,\\ B&=2. \end{aligned}$$
Let $x=1$: $$\begin{aligned} 3(1)-5&=A(1-3),\\ -2&=-2A,\\ A&=1. \end{aligned}$$
Therefore, $$\dfrac{3x-5}{x^2-4x+3} = \dfrac{1}{x-1} + \dfrac{2}{x-3}.$$

EXAMPLE 6 (Repeated Linear Factor)

Decompose $$\dfrac{5x+1}{(x-2)^2}$$ into partial fractions.

Solution:

Since the denominator has a repeated linear factor, write $$\dfrac{5x+1}{(x-2)^2} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2}.$$
Multiply both sides by $(x-2)^2$: $$5x+1=A(x-2)+B.$$
Compare coefficients: $$\begin{aligned} 5x+1&=Ax-2A+B,\\ A&=5,\\ -2A+B&=1. \end{aligned}$$
Substitute $A=5$: $$\begin{aligned} -2(5)+B&=1,\\ B&=11. \end{aligned}$$
Therefore, $$\dfrac{5x+1}{(x-2)^2} = \dfrac{5}{x-2} + \dfrac{11}{(x-2)^2}.$$

EXAMPLE 7 (Irreducible Quadratic Factor)

Decompose $$\dfrac{2x+3}{(x-1)(x^2+4)}$$ into partial fractions.

Solution:

Since $x^2+4$ is irreducible over $\mathbb{R}$, write $$\dfrac{2x+3}{(x-1)(x^2+4)} = \dfrac{A}{x-1} + \dfrac{Bx+C}{x^2+4}.$$
Multiply both sides by $(x-1)(x^2+4)$: $$2x+3=A(x^2+4)+(Bx+C)(x-1).$$
Expand and collect terms: $$\begin{aligned} 2x+3&=Ax^2+4A+Bx^2-Bx+Cx-C,\\ &=(A+B)x^2+(C-B)x+(4A-C). \end{aligned}$$
Compare coefficients: $$\begin{aligned} A+B&=0,\\ C-B&=2,\\ 4A-C&=3. \end{aligned}$$
Solving gives $$A=1,\qquad B=-1,\qquad C=1.$$
Therefore, $$\dfrac{2x+3}{(x-1)(x^2+4)} = \dfrac{1}{x-1} + \dfrac{-x+1}{x^2+4}.$$