Question

What is an oblique asymptote?
Will medal and fan!

Answer 1

@zepdrix

Answer 2

A rational function has at most one horizontal asymptote or oblique (slant) asymptote, and possibly many vertical asymptotes. The degree of the numerator and degree of the denominator determine whether or not there are any horizontal or oblique asymptotes.

Answer 3

thx

Answer 4

Does this graph have a oblique asymptote: \frac{x^2+6-9}{x-2} @cupcakeohla

Answer 5

\[\frac{x^2+6-9}{x-2}\]

Answer 6

1 sec

Answer 7

Okay.

Answer 8

mabey this will help ?

Answer 9

Predict if any asymptotes or holes are present in the graph of each rational function. Use a graphing calculator to draw the graph and verify your prediction. In Part A, we'll graph y = x/(x2 – 9).
Let's take a moment to learn about rational functions with a numerator degree less than the denominator degree.
There is a horizontal asymptote at y = 0, along the x-axis.
We also have vertical asymptotes at the non-permissible values of the rational function.
We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.
Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero.
Write the equation x2 – 9 = 0.
Factor the difference of squares to get (x + 3)(x – 3) = 0.
The roots of the equation are x = -3 and x = 3.
We have vertical asymptotes at x = -3 and x = 3.
Now graph the rational function using technology. The asymptotes we predicted exist on the graph. When typing this into your graphing calculator, make sure the denominator is in brackets.
Is it an error that the graph is crossing the horizontal asymptote?
No. The horizontal asymptote is just a guide for the end behaviour of the graph. It's fine if intermediate points exist on the asymptote.
Contrast this with vertical asymptotes, where the graph is never allowed to cross the asymptote.
In Part B, we'll graph y = (x + 2)/(x2 + 1).
We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.
Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero.
Write the equation x2 + 1 = 0.
Subtract 1 from both sides of the equation to get x2 = -1.
Square root both sides to get x = root(-1).
root(-1) is undefined. There are no vertical asymptotes.
Now graph the rational function using technology. The asymptotes we predicted exist on the graph. When typing this into your graphing calculator, make sure the numerator and denominator are in brackets.
In Part C we'll graph y = (x + 4)/(x2 – 16).
We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.
Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero.
Write the equation x2 – 16 = 0.
Add 16 to both sides of the equation.
Square root both sides to get x = root(16).
This gives us x = ±4.
Based on this result, we should have vertical asymptotes at x = -4 and x = 4.
Now graph the rational function using technology. Use brackets for the numerator and denominator.
The graph unexpectedly crosses the vertical asymptote at x = -4. What's going on here?
Factor the denominator and simplify. This gives us 1/(x – 4).
There is information loss when we cancel x + 4 from the numerator and denominator. As far as the graph is concerned, there is only one non-permissible value, +4.
However, we know for a fact that x = -4 is a non-permissible value! How do we indicate this on the graph?
Use an open circle to indicate that when x = -4, the graph does not exist.
In words, we can say: "The graph is discontinuous at x = -4", or "The graph has a hole at x = -4".
In Part D, we'll graph y = (x2 – x – 2)/(x3 – x2 – 2x).
We'll begin by finding the horizontal asymptote. Since the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote at y = 0.
Now we'll find the vertical asymptotes. Vertical asymptotes occur at the values of x that make the denominator become zero.
Write the equation x3 – x2 – 2x = 0.
Factor out the x.
Now factor the trinomial.
The roots of the equation are -1, 0, and 2, in ascending numerical order.
There should be vertical asymptotes at x = -1, x = 0, and x = 2.
Now graph the rational function using technology. Use brackets for the numerator and denominator.
The asymptotes at x = -1 and x = 2 are ignored.
Factor the rational expression and cancel identical factors. This gives us 1/x.
As far as the graph is concerned, only zero is a non-permissible value. This is the result of information loss from canceling identical factors.
However, we know for a fact that -1 and 2 are non-permissible values. Draw holes in the graph where x = -1 and x = 2 to indicate this.

Answer 10

Wow that's a lot of words 0_o

Answer 11

Yeah… maybe a step by step and not a copy-and-paste from a textbook?

Answer 12

Oblique asymptote = diagonal or slant asymptote.

Remember when you do Synthetic Division or Polynomial Long Division,
you end up with a `quotient` and a `remainder`.

The quotient gives you the equation of your oblique asymptote.
I think.

I'm a little rusty on these :) I better check before I go spouting off nonsense.

Answer 13

Are you familiar with either method?
Synthetic Division?
Polynomial Long Division?

Preference?

Answer 14

this

Answer 15

idk sorry

Answer 16

It's fine c: we got it

Answer 17

Yes, synthetic division. I divided them and got x + 14 (remainder 19).