How do you find vertical, horizontal and oblique asymptotes for #(x^2-x+1)/(x-3)#?

Question

Elias Ackerman · Accepted Answer

The vertical asymptote is #x=3#
No horizontal asymptote.
The oblique asymptote is #y=x+2# As you cannot divide by #0#, #x!=3# Consequently, The vertical asymptote is #x=3# As the degree of the numerator is #># than the degree of the denominator, we expect an oblique asymptote. Let's divide something long. #color(white)(aaaa)##x^2-x+1##color(white)(aaaa)##∣##x-3# #color(white)(aaaa)##x^2-3x##color(white)(aaaaaaa)##∣##x+2# #color(white)(aaaaa)##0+2x+1# #color(white)(aaaaaaa)##+2x-6# #color(white)(aaaaaaa)##+0+7# Consequently, #(x^2-x+1)/(x-3)=x+2+7/(x-3)# The oblique asymptote is #y=x+2# There is no horizontal asymptote. When #x->+-oo#, #f(x)->+-oo# graph{(y-x-2)=0 [-41.1, 41.1, -20.56, 20.56]} or y-(x^2-x+1)/(x-3)

Madelyn Anderson · Answer

undefined To find vertical, horizontal, and oblique asymptotes for the function $\frac{x^2-x+1}{x-3}$:

1. Vertical asymptotes: Set the denominator equal to zero and solve for $x$. The values of $x$ obtained will give the vertical asymptotes.
$$x - 3 = 0$$
$$x = 3$$

2. Horizontal asymptotes: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
In this case, the degree of the numerator (2) is greater than the degree of the denominator (1), so there is no horizontal asymptote.

3. Oblique asymptotes: If the degree of the numerator is exactly one more than the degree of the denominator, divide the numerator by the denominator using long division or synthetic division to find the equation of the oblique asymptote.
$$x^2 - x + 1 = (x - 3)(x + 2) - 5$$
So, the oblique asymptote is $y = x + 2$.