For the following function f(x) =2x^2-3x-9/x^2-2x-3. Find the vertical, horizontal and oblique asymptote and hole in the graph (if any). Show all steps.

Question

nottim · Answer

You MUST have a note on this.

nottim · Answer

To find the...asymptote of something sub infinity in for x (that's not a joke; that's for eitehr vertical or horizontal; pretty sure horizontal)

nottim · Answer

Since oblique is just a line...

nottim · Answer

diagonal one..

anonymous · Answer

Okay.. can you be more specific please?

nottim · Answer

I just don't know. That's the truth. I was hoping that maybe you had a textbook or note for this.

nottim · Answer

I remember another required some table of values, maybe vertical asymptote.

nottim · Answer

i...just cant remember. I suggest trying Khan Academy or just using google to look up the meethods...

nottim · Answer

If you really don't have a textbook or note on this.

nottim · Answer

hey?

anonymous · Answer

Okay, i'm checking. Thank you.

nottim · Answer

yeah. sorry that I cant tell you directly; but I haven't memorized the info myself.

campbell_st · Answer

factorising

$$((2x +3)(x -3))/(x+1)(x-3)$$

gives

$$(2x+3)/(x+1)$$

a vertical asymptote exists at x = -1

a point of discontinuity exists at x =3, you can find the y value for the ordered pair
this is because the original equation before simplifying has the restriction

$$x \neq 3$$

use synthetic or polynomial division to find the oblique(horizontal) asymptote

it exists at y = 2

anonymous · Answer

since the factor of (x - 3) cancelled, that is where you have a hole in your graph.   It is a point of discontinuity... removable one at that.