How do you find the vertical, horizontal or slant asymptotes for #(4x^2+5)/( x^2-1)#?

Question

Christian Antunez · Accepted Answer

Vertical asymptotes: #x = 1# and #x = -1#
Horizontal asymptote: #y = 4# Ok, let's start with the vertical asymptotes. You know that a function is not defined when a denominator equals #0#. In this case, that would be when #x^2 - 1 = 0#. Using basic equation rules, can can change that into #x^2 = 1#. Then, we take the square root of both sides, making sure to note the positive and negative possibilities. This gives us: #x = 1# or #x = -1# I'm not sure how complicated your issues will get, but at this point, either a removable discontinuity or a vertical asymptote could occur. If you don't know what a removable discontinuity is, you can probably skip the following section. Discontinuities that can be removed: Try factoring both the top and the bottom to see if you have a removable discontinuity; if any of the factors cancel, that's when you have a removable discontinuity. Here's an example: #(25 - x^2)/(5-x)# This contributes to: #((5 + x)(5-x))/(5-x)# Now you notice that the #(5-x)#'s cancel leaving: #5 + x# or #x + 5# (if you prefer the #x# first) This makes a line, which is pretty easy to graph. However, because #5-x = 0# when #x# is #5#, it cannot be in the domain. Therefore, you are left with the line #y = x + 5# with a domain of all real numbers except for #5#. At that point, there is simply a hole in the graph notated by an empty circle where the point would be if it were in the domain (in my example, that point would be at #(5, 10)#). However, the top part of your problem doesn't even factor, so there cannot be a removable discontinuity giving you vertical asymptotes at #x = 1# and #x = -1# Final Breakable Interruptions Let's move on to the asymptotes that are horizontal and oblique, or slant as you put it. This is done by doing the following: First take the highest powered variables on the top and the bottom and remove the rest. In your case this would be #(4x^2)/(x^2)# We can do this because horizontal and oblique asymptotes are when #x# is really big, meaning that only this term is going to be relevant. For example, is #1,000,001# that different from #999,998#? Not really. Now, determine which power is greater; #x^3# is bigger than #x^2# no matter how big or small the coefficients (the numbers in front of the variables) are. If the top is larger, simplify! Your oblique asymptote is what you end up with. If the bottom is bigger: You got off easy. The horizontal asymptote is just #y = 0# If they are equal (like your problem): Divide the coefficients (in your case it's #4/1#, which is just #4#) and the horizontal asymptote is #y = #whatever you got. That means that your horizontal asymptote is #y = 4#. I hope this was useful. Moore, Jonathan 'JMoney'

Benjamin Achtenberg · Answer

undefined To find the vertical asymptotes of the function (4x^2 + 5)/(x^2 - 1), we identify the values of x where the denominator equals zero, excluding any roots that cancel out with the numerator.

To find horizontal asymptotes, we compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0. If the degrees are equal, the horizontal asymptote is given by the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find slant (oblique) asymptotes, we perform polynomial division to see if the function can be written as a polynomial plus a proper rational function. If the degree of the polynomial obtained from the division is one greater than the degree of the denominator, there is a slant asymptote.

To summarize:

1. Vertical asymptotes: Set the denominator equal to zero and solve for x, excluding any roots that cancel out with the numerator.
2. Horizontal asymptotes: Compare the degrees of the numerator and denominator.
3. Slant asymptotes: Perform polynomial division to see if the function can be expressed as a polynomial plus a proper rational function.

For the given function (4x^2 + 5)/(x^2 - 1):
1. Vertical asymptotes: Set x^2 - 1 = 0 ⇒ x^2 = 1 ⇒ x = ±1.
2. Horizontal asymptotes: The degrees of the numerator and denominator are equal, so we look at the ratio of leading coefficients: 4/1 = 4. Therefore, there is a horizontal asymptote at y = 4.
3. There is no slant asymptote because the degree of the numerator (2) is not one greater than the degree of the denominator (2).