Rational functions in the form $ y = {P(x) \over D(x)}$

Improper rational functions

$$ y = {P(x) \over D(x)} $$

$$ y = {2x \over 2x - 1} $$

\begin{align} {\text{Polynomial} \over \text{Divisor}} & = \text{Quotient} + {\text{Remainder} \over \text{Divisor}} \\ \\ y & = {2x \over 2x - 1} \\ & = 1 + {1 \over 2x - 1} \end{align}

Rational function 1: $y = {a \over x + b}$

$$ y = {a \over x + b} $$

In this form, the features of the graph are:

• Horizontal asymptote: $y = 0$
• Vertical asymptote: $x = - b$
• Stationary point(s) may not exist - let ${dy \over dx} = 0$ and (try to) solve for coordinates of stationary points

Example

Consider the graph of $y = {1 \over x + 2}$:

\begin{align} y & = {1 \over x + 2} \\ \\ \text{Horizontal asymptote: } & y = 0 \\ \text{Vertical asymptote: } & x = -2 \\ \\ \text{Let } & x = 0, \\ y & = {1 \over 0 + 2} \\ y & = 0.5 \\ \implies y-\text{int} & \text{ercept is } 0.5 \\ \\ \text{No } x \text{ -intercepts } & \text{since } y = 0 \text{ is the asymptote} \\ \\ y & = {1 \over x + 2} = (x + 2)^{-1} \\ \\ {dy \over dx} & = (-1)(x + 2)^{-2} (1) \phantom{000} [\text{Chain rule}] \\ & = -{1 \over (x + 2)^2} \\ \\ \text{For all real va} \text{lues of } & x \text{ (excluding } x = -2), \\ (x + 2)^2 & > 0 \\ \implies -{1 \over (x + 2)^2 } & < 0 \\ {dy \over dx} & < 0 \\ \\ \therefore \text{No stationary} & \text{ points (since } {dy \over dx} \ne 0 \text{ for any values of } x) \end{align}

Rational function 2: $y = p + {a \over x + b}$

$$ y = p + {a \over x + b} $$

In this form, the features of the graph are:

• Horizontal asymptote: $y = p$
• Vertical asymptote: $x = - b$
• Stationary point(s) may not exist - let ${dy \over dx} = 0$ and (try to) solve for coordinates of stationary points

Example

Consider the graph of $y = 1 + {1 \over 2x - 1}$:

\begin{align} \text{Horizontal asymptote: } & y = 1 \\ \text{Vertical asymptote: } & x = 0.5 \\ \\ \text{Let } & x = 0, \\ y & = 1 + {1 \over 2(0) - 1} \\ y & = 0 \\ \implies x/y-\text{int} & \text{ercept is } 0 \\ \\ y & = 1 + {1 \over 2x - 1} = 1 + (2x - 1)^{-1} \\ \\ {dy \over dx} & = 0 + (-1)(2x - 1)^{-2} (2) \\ & = -{2 \over (2x - 1)^2} \\ \\ \text{For all real va} \text{lues of } & x \text{ (excluding } x = 0.5), \\ (2x - 1)^2 & > 0 \\ \implies -{2 \over (2x - 1)^2 } & < 0 \\ {dy \over dx} & < 0 \\ \\ \therefore \text{No stationary} & \text{ points (since } {dy \over dx} \ne 0 \text{ for any values of } x) \end{align}

Rational function 3: $y = px + q + {a \over x + b}$

$$ y = px + q + {a \over x + b} $$

In this form, the features of the graph are:

• Oblique asymptote: $y = px + q$
• Vertical asymptote: $x = - b$
• Stationary point(s) may not exist - let ${dy \over dx} = 0$ and (try to) solve for coordinates of stationary points

Example

Consider the graph of $y = -x + 1 - {1 \over x - 1}$:

\begin{align} y & = -x + 1 - {1 \over x - 1} \\ \\ \text{Oblique asymptote: } & y = -x + 1 \\ \text{Vertical asymptote: } & x = 1 \\ \\ \text{Let } & x = 0, \\ y & = -(0) + 1 - {1 \over 0 - 1} \\ y & = 2 \\ \implies y-\text{int} & \text{ercept is } 2 \\ \\ y & = -x + 1 - {1 \over x - 1} \\ & = -x + 1 - (x - 1)^{-1} \\ \\ {dy \over dx} & = -1 + 0 - (-1)(x - 1)^{-2} (1) \\ & = - 1 + {1 \over (x - 1)^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = -1 + {1 \over (x - 1)^2} \\ 1 & = {1 \over (x - 1)^2} \\ (x - 1)^2 & = 1 \\ x - 1 & = \pm \sqrt{1} \\ x - 1 & = \pm 1 \\ x & = 1 + 1 \text{ or } - 1 + 1 \\ x & = 2 \text{ or } 0 \\ \\ \implies (0, 2) & \text{ is a stationary point} \\ \\ \text{When } & x = 2, \\ y & = -(2) + 1 - {1 \over 2 - 1} \\ y & = -2 \\ \implies (2, -2) & \text{ is a stationary point} \end{align}