Oblique Asymptote

[monopolyman]

I was working on my homework and came across this problem which really stumped me:

Instructions: Give the x and y intercepts as well as the horizontal and vertical asymptotes. Sketch a graph of the function.

f(x) = (x^2 -x)/(x + 2)

Here is what I have so far, though I am unsure if I am doing this correctly or not:

x-int: "x^2 - x = 0"
y-int: -1/2 ----- f(x) = (x^2 -x) /(x + 2)----- coefficient in front of the x is -1 as in -(1)x
v.a.: x = -2 ----- f(x) = (x^2) -x) / (x + 2) ----- x + 2 = 0.....x = -2
h.a.: No horizontal asymptote, as it is oblique.

As it is oblique, is the vertical asymptote I listed even correct? " x = " implies a straight vertical line, not a diagonal one as is with an oblique graph.

I have no clue how to find the x-intercept for this equation.... "x^2 - x = 0" cannot be solved, to my knowledge.

I would greatly appreciate any help you can provide.


EDIT: I should be able to graph it once I have the above information properly filled out.

 

[srmichael]

Do you see that \(\displaystyle x^2-x\) can be factored to \(\displaystyle x(x-1)\). Do this, then you can solve for the x-intercept.

The y-intercept is when x=0. The y-intercept is NOT -1/2.

Vertical Asymptote is correct.

And, yes, there is no horizontal asymptote. There is what is called a slant asymptote.

 

[monopolyman]

Do you see that \(\displaystyle x^2-x\) can be factored to \(\displaystyle x(x-1)\). Do this, then you can solve for the x-intercept.

The y-intercept is when x=0. The y-intercept is NOT -1/2.

Vertical Asymptote is correct.

And, yes, there is no horizontal asymptote. There is what is called a slant asymptote.

Ohhhh ok, thank you very much!

So x = 1 right?

y = ((1)^2 -(1)) / ((1) + 2)

y = 0 ? Or am I doing this wrong?

 

[srmichael]

So x = 1 right?

y = ((1)^2 -(1)) / ((1) + 2)

y = 0 ? Or am I doing this wrong?

Almost. X-intercept is x=1 and x=0. You have to set both factors equal to zero.

Yes, the y-intercept is 0, but you found it incorrectly. You plugged in x=1 into the equation. You need to plug in x=0 in to the equation. All y-intercept points are of the nature (0,y), thus x=0. So \(\displaystyle y=\frac{0^2-0}{0+2} = \frac{0}{2} = 0\).

 

[monopolyman]

Got it, thank you very much!

 

[monopolyman]

Sorry, I'm having trouble graphing this now. I thought it was an oblique (slant) asymptote...How do you know how steep it is with only x = -2 to go off of? Also, the x/y ints are all on the same axis, which is confusing me for some reason. It's been about a week since we went over this in class and I am afraid I've forgotten the majority of it...Any suggestions, help, or advice to offer?

EDIT: I've also plugged this into Wolfram Alpha to see what it to say and it said it's asymptotic to x - 3....do I have my 4 steps wrong:

f(x) = (x^2 - x) / (x + 2)

1) x-int: (0, 0), (1, 0)
2) y-int: (0, 0)
3) V.A.: x = -2
4) H.A.: None

 

[srmichael]

View attachment 1465
Sorry, I'm having trouble graphing this now. I thought it was an oblique (slant) asymptote...How do you know how steep it is with only x = -2 to go off of? Also, the x/y ints are all on the same axis, which is confusing me for some reason. It's been about a week since we went over this in class and I am afraid I've forgotten the majority of it...Any suggestions, help, or advice to offer?

EDIT: I've also plugged this into Wolfram Alpha to see what it to say and it said it's asymptotic to x - 3....do I have my 4 steps wrong:

f(x) = (x^2 - x) / (x + 2)

1) x-int: (0, 0), (1, 0)
2) y-int: (0, 0)
3) V.A.: x = -2
4) H.A.: None

You need to do long division. Do you know how to do long division of polynomials? If you do \(\displaystyle \frac{x^2-x}{x+2}\) you will see that your quotient will be \(\displaystyle x-3\) which is your slant asymptote.

 

[monopolyman]

You need to do long division. Do you know how to do long division of polynomials? If you do \(\displaystyle \frac{x^2-x}{x+2}\) you will see that your quotient will be \(\displaystyle x-3\) which is your slant asymptote.

So what's up with the vertical asymptote showing x = -2?

 

[lookagain]

[/tex] ...you will see that your quotient will be \(\displaystyle x-3\) \(\displaystyle from \ which \ y = x - 3\) is your slant asymptote.

Oblique asymptotes (as well as horizontal and vertical ones) are lines,

so there is an \(\displaystyle equation\) associated with them.