Question

let $f(x)=\frac{3x^{2}-16x + 5}{3x^{2}-8x + 4}$
this function has (enter answers separated by commas if more than one):

1. vertical asymptotes at $x=$
2. horizontal asymptote at $y=$

question help: message instructor

Explanation:

Step1: Factor numerator and denominator

Factor \(3x^{2}-16x + 5=(3x - 1)(x - 5)\)
Factor \(3x^{2}-8x + 4=(3x - 2)(x - 2)\)
So \(f(x)=\frac{(3x - 1)(x - 5)}{(3x - 2)(x - 2)}\)

Step2: Find vertical asymptotes

Vertical asymptotes occur where denominator is \(0\) (and numerator is not \(0\)).
Set \(3x-2 = 0\), then \(x=\frac{2}{3}\)
Set \(x - 2=0\), then \(x = 2\)

Step3: Find horizontal asymptote

Since degree of numerator \(n = 2\) and degree of denominator \(m=2\) (where \(n=m\))
The horizontal asymptote \(y=\frac{a_{n}}{b_{m}}\), where \(a_{n}\) is leading coefficient of numerator and \(b_{m}\) is leading coefficient of denominator.
Here \(a_{n}=3\), \(b_{m}=3\), so \(y = 1\)

Answer:

1. \(x=\frac{2}{3},2\)
2. \(y = 1\)