How do you find vertical, horizontal and oblique asymptotes for #(x^2 - 5x + 6)/ (x - 3)#?

Question

Madelyn Anderson · Accepted Answer

Remember: You cannot have three asymptotes at the same time. If the Horizontal Asymptote exists, the Oblique Asymptote doesn't exist. Also, #color (red) (H.A)# #color (red) (follow)# #color (red) (three)# #color (red) (procedures).# Let's say #color (red)n# = highest degree of the numerator and #color (blue)m# = highest degree of the denominator,#color (violet) (if)#:
#color (red)n color (green)< color (blue) m#, #color (red) (H.A => y = 0)#
#color (red)n color (green)= color (blue) m#, #color (red) (H.A => y = a/b)#
#color (red)n color (green)> color (blue) m#, #color (red) (H.A) # #color (red) (doesn't)# #color (red) (EE)#

Here, #(x^2 - 5x + 6)/(x-3)#

#V.A: x-3=0 => x = 3#
#O.A: y=x-2#

Please, take a look at the picture.

The oblique/slant asymptote is found by dividing the numerator by the denominator (long division.)

Notice that I did not do the long division in the way some people excepted me to. I always use the "French" way because I've never understood the English way, also I'm a francophone :) but it is the same answer.

Hope this helps :)

Caleb Alexander · Answer

undefined To find the vertical asymptote of a rational function, set the denominator equal to zero and solve for $x$.

$x - 3 = 0$

$x = 3$

So, there is a vertical asymptote at $x = 3$.

To find horizontal asymptotes, analyze the behavior of the function as $x$ approaches positive or negative infinity.

As $x$ approaches positive or negative infinity, the function behaves like the ratio of the leading terms of the numerator and the denominator.

For the given function $\frac{x^2 - 5x + 6}{x - 3}$:

- The leading term of the numerator is $x^2$.
- The leading term of the denominator is $x$.

So, as $x$ approaches positive or negative infinity, the function behaves like $\frac{x^2}{x}$, which simplifies to $x$.

Thus, there is a horizontal asymptote at $y = x$.

There are no oblique asymptotes for this function.