Let f(x)= (3x^2-2x+1)/(x-1). What are the types of asymptotes of f(x)? What are the equations of all the asymptotes of f(x)?
The types of asymptote are vertical and oblique?
x=1?
Is there something wrong I did
asymptotes are indeed vertical (x=1) and oblique (y=3x).
y=3x because for lage x, the fraction just looks like 3x^2/x = 3x.
No, there is nothing wrong with what you did. Let's go through the process of finding the types of asymptotes and their equations for the given function f(x) = (3x^2 - 2x + 1)/(x - 1).
To determine the types of asymptotes, we need to look at the behavior of the function as x approaches certain values. We consider two cases:
1. Vertical Asymptotes: These occur when the function approaches positive or negative infinity as x approaches a particular value. To find the vertical asymptotes, we look for values of x that would make the denominator of the function equal to zero.
In this case, for f(x) = (3x^2 - 2x + 1)/(x - 1), we find that the denominator (x - 1) would be equal to zero when x = 1. Therefore, x = 1 is a vertical asymptote.
2. Oblique (Slant) Asymptotes: These occur when the function approaches a non-horizontal line as x approaches positive or negative infinity. To find oblique asymptotes, we need to divide the polynomial numerator (3x^2 - 2x + 1) by the polynomial denominator (x - 1).
Performing long division:
____________________
x - 1 | 3x^2 - 2x + 1
-(-3x^2 + 3x)
________________
1 - x + 1
- (1 - x)
______________
2
The quotient of the division is 3 + (2/(x - 1)), which represents the linear term and remainder.
Since the quotient is not a polynomial and includes (2/(x - 1)), an expression with a variable in the denominator, we have an oblique asymptote.
Therefore, the types of asymptotes for f(x) are one vertical asymptote (x = 1) and one oblique asymptote.
The equation of the vertical asymptote is x = 1.
The equation of the oblique asymptote is y = 3 + (2/(x - 1)).