How do you identify all asymptotes for #f(x)=(x^2-3x+2)/x#?

Question

Serenity Jones · Accepted Answer

See below.

Vertical asymptotes occur where the function is undefined, for:

#(x^2-3x+2)/x#

This is undefined for #x=0# ( division by zero )

Vertical asymptote is the line: #x=0#

Notice that the degree of the numerator is greater than the degree of the denominator. In this case there is an oblique asymptote. This will be a line of the form #y=mx+b#. To find this line, we divide the numerator by the denominator. We only need to divide until we have the equation of a line.

#:.#

Dividing by #x#:

#(x^2/x-3x/x+2/x)/(x/x)=x-3+2/x#

So the oblique asymptote is the line:

#color(blue)(y=x-3)#

The graph confirms these findings:

Liam Brennan · Answer

undefined To identify all asymptotes for $ f(x) = \frac{x^2 - 3x + 2}{x} $, you need to consider two types of asymptotes:

1. Vertical asymptotes: Vertical asymptotes occur where the function approaches positive or negative infinity as x approaches a certain value. These occur where the denominator of the function becomes zero but the numerator doesn't. In this case, the vertical asymptote occurs at $ x = 0 $.

2. Horizontal asymptotes: Horizontal asymptotes occur when the function approaches a constant value as x approaches positive or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator of the function. In this case, since the degree of the numerator (which is 2) is greater than the degree of the denominator (which is 1), there are no horizontal asymptotes.

Therefore, for the function $ f(x) = \frac{x^2 - 3x + 2}{x} $, the vertical asymptote is $ x = 0 $, and there are no horizontal asymptotes.