Given the function `f(x)=(x^2-x+8)/(x^2-4x+3)` :

(1) Find the vertical asymptotes:

A rational function has a vertical asymptote at any value of x that causes the denominator to be equal to zero while the numerator is nonzero. To find the zeros of the numerator and denominator we can factor:

`(x^2-x+8)/(x^2-4x+3)=(x^2-x+8)/((x-3)(x-1))`

There...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Given the function `f(x)=(x^2-x+8)/(x^2-4x+3)` :

(1) Find the vertical asymptotes:

A rational function has a vertical asymptote at any value of x that causes the denominator to be equal to zero while the numerator is nonzero. To find the zeros of the numerator and denominator we can factor:

`(x^2-x+8)/(x^2-4x+3)=(x^2-x+8)/((x-3)(x-1))`

There are vertical asymptotes at x=1 and x=3.

(2) Since the degree of the numerator is the same as the degree of the denominator, there is a horizontal asymptote at `y=a_n/b_m` where `a_n,b_m` are the leading coefficients of the numerator and denominator respectively when written in standard form.

The horizontal asymptote is y=1.

(3) We are asked to find four ordered pairs that are on the graph of the function.

`f(-4)=((-4)^2-(-4)+8)/((-4)^2-4(-4)+3)=28/35=4/5`

`f(-3)=((-3)^2-(-3)+8)/((-3)^2-4(-3)+3)=20/24=5/6`

`f(-2)=((-2)^2-(-2)+8)/((-2)^2-4(-2)+3)=14/15`

`f(-1)=((-1)^2-(-1)+8)/((-1)^2-4(-1)+3)=10/8=5/4`

`f(0)=8/3` (This is the y-intercept.)

`f(2)=((2)^2-(2)+8)/((2)^2-4(2)+3)=10/(-1)=-10`

`f(4)=((4)^2-(4)+8)/((4)^2-4(4)+3)=20/3`

So some points on the graph include:

`(-4,4/5),(-3,5/6),(-2,14/15),(-1,5/4),(0,8/3),(2,-10)`

`(4,20/3),(5,7/2),(6,38/15)`

The graph: