Chapter 4, Problem 49RE

To determine

To design: The rational function.

Answer to Problem 49RE

The rational function is $f(x)=\frac{2(x-2)(x-3/2)(x-1)}{(x+2)(x-3)}$ .

Explanation of Solution

Given information:

The characteristics are: three real zeros , one multiplicity 2; y -intercept 1; vertical asymptotes $x=-2$ and $x=3$ ; oblique asymptote $y=2x+1$ .

Formula used:

Vertical Asymptote: A rational function $R(x)=\frac{p(x)}{q(x)}$ , in lowest terms, will have a

asymptote $x=r$ if $r$ is a real zero of the denominator q .

Oblique Asymptote : it occur when the degree of the denominator of a rational function is one less than the degree of the numerator. The asymptote will be along line

  $y=mx+b$ where $m\ne 0$

Calculation:

We have ,

Real zeros: $3$ , one with multiplicity 2

Vertical Asymptote: $x=-2,3$

Oblique Asymptote: $y=2x+1$

y -intercept: 1

The general form is

  $=f(x)=\frac{P(x)}{Q(x)}$ with $d^2-4e<0$

Where P(x) and Q(x) are polynomial functions of x and Q is not the zero function.

Using vertical asymptote we get,

  $=(x+2)(x-3)$ (i)

And Oblique asymptote:

  $=y=2x+1;m=2;c=1$ (ii)

Therefore,

Denominator : $(x+2)(x-3)$ (iii)

Numerator: $=a(x-b)(x-c)(x-d)$

  $=2(x-2)(x-c)(x-d)$ (iv)

By (iii) and (iv) we get,

  $=f(x)=\frac{2(x-2)(x-c)(x-d)}{(x+2)(x-3)}$

y-intercept :

  $=f(0)=1$

  $=\frac{2(x-2)(x-c)(x-d)}{(x+2)(x-3)}=1$

  $\begin{array}{l}=\frac{2(0-2)(2-c)(0-d)}{(0+2)(0-3)}=1\\ =\frac{2(-2)(-c)(-d)}{(2)(-3)}=1\\ =\frac{4cd}{6}=1\\ =cd=\frac{3}{2}\end{array}$

It is not unique.

Hence,

  $f(x)=\frac{2(x-2)(x-3/2)(x-1)}{(x+2)(x-3)}$