Equation for horizontal and oblique asymptotes

[sunwers]

I have a question about identifying the horizontal and oblique asymtotes

In this chapter the notes states

1. If n < m, the horizontal line y=0 (the x-axis) is the horizontal asymptote for f.

(-4)/(x+9) y=0

2. If n = m, the horizontal line y=(a^n/b^m) is the horizontal asymptote for f.

(x^2+9x-5)/(-x^2+36)
(1x^2)/(-1x^2) = Y=-1

3. If n = m + 1, the line y = g(x) is an oblique asymptote for f, where g is the quotient polynomial obtained by dividing p by q (the remainder polynomial is irrelevant).

(8x^2+28x+20)/(-2x-9)

degrees are n = m + 1

sorry not sure how to show my division problem so I inserted the image
y=-4x+64


4. If n > m + 1, there is no straight-line or oblique asymptote for f.

This one is what I'm not sure of. I'm thinking if the remainder show 0 then it maybe none but I don't know.

Thanks,
Sunwers

 

[lookagain]

I have a question about identifying the horizontal and oblique asymtotes

In this chapter the notes states

1. If n < m, the horizontal line y=0 (the x-axis) is the horizontal asymptote for f.

y = (-4)/(x + 9) \(\displaystyle \ \ \ \) y=0

2. If n = m, the horizontal line y=(a^n/b^m) is the horizontal asymptote for f.

y = (x^2 + 9x - 5)/(-x^2 + 36)
(1x^2)/(-1x^2) = y = -1

3. If n = m + 1, the line y = g(x) is an oblique asymptote for f, where g is the quotient polynomial obtained by dividing p by q (the remainder polynomial is irrelevant).

y = (8x^2 + 28x + 20)/(-2x - 9)

degrees are n = m + 1

sorry not sure how to show my division problem so I inserted the image
View attachment 3395 y = -4x + 64 \(\displaystyle \ \ \) <---- This must be a typo, because in the box you have it as "y = -4x + 4."


4. If n > m + 1, there is no > > > straight-line or oblique asymptote < < < \(\displaystyle \ \ \ \) <--- What do you mean by this?
All of the asymptotes under consideration in this narrowed context *are* straight lines. I think you meant to type
"horizontal lines or oblique lines."



for f.

This one is what I'm not sure of. I'm thinking if the remainder show 0 then it maybe none but I don't know. \(\displaystyle \ \ \ \) If I understand you correctly,
then I will state that the remainder is irrelevant if the degree of the numerator in a rational function is more than one greater than the degree
of the denominator.

Without "y =" or "f(x) =," for example, in front of those rational expressions in x, they are not equations of curves at all.

 

[sunwers]

Without "y =" or "f(x) =," for example, in front of those rational expressions in x, they are not equations of curves at all.

Number 4 is in the notes in my text "If n > m + 1, there is no straight-line horizontal or oblique asymptote for f."

I'm thinking this equation (x^3-5)/(x+4) would be no horizontal and oblique asymptotes as the answer for that problem. (a^n)/(b^m) So is if the n degree of the leading coefficient is greater than 1 degree of the m degree leading coefficient... does this mean no horizontal and oblique asymptotes?

 

[lookagain]

Number 4 is in the notes in my text "If n > m + 1, there is no straight-line horizontal or oblique asymptote for f.


"I'm thinking this equation (x^3-5)/(x+4) \(\displaystyle \ \ \ \) That's *not* an equation! y = (x^3 - 5)/(x + 4) is an equation. It has an equals sign.


would be no horizontal and oblique asymptotes as the answer for that problem. (a^n)/(b^m) So is if the n degree of the leading coefficient is greater than 1 degree of the m degree leading coefficient... does this mean no horizontal and oblique asymptotes?

For y = (x^3 - 5)/(x + 4), it is true that it has neither horizontal nor oblique asymptotes.

 

[sunwers]

For y = (x^3 - 5)/(x + 4), it is true that it has neither horizontal nor oblique asymptotes.

Ok thanks for your help I'm sorry I didn't put the y= in the equation will do the next time.

 

[HallsofIvy]

You should not have to rely on memorized rules. Dividing \(\displaystyle x^3- 5\) by \(\displaystyle x+ 4\) gives \(\displaystyle x^2- 4x+ 5- \dfrac{25}{x+ 4}\). As x gets larger and larger, \(\displaystyle \dfrac{25}{x+ 4}\)= goes to 0. Do you see why there is no linear asymptote?