Horizontal Asymptotes

[nycmathguy]

Section 2.6
10, 12, 14, 16

Q. 10

The numerator is 1, which can be expressed as 1•x^0. The degree of the numerator is smaller than the degree of the denominator. The horizontal asymptote is the line y = 0.

Q. 12

The degree of the numerator = degree of denominator. The horizontal asymptote is -7/2 for this question.

Q. 14

The degree of the numerator > degree of denominator. So, no horizontal asymptote.

Q. 16

The degree of the numerator = degree of the denominator. So, the horizontal asymptote is
-4/1 or simply -4.

You say?

 

[MathLover1]

Q. 10

The horizontal asymptote is the line y = 0-> correct
The vertical asymptote is the line x =2

12.
Vertical: x=- 3/2
Horizontal: y=-7/2 -> you did it correct

Q. 14
no horizontal asymptote=> correct
Vertical: x= -2

Q. 16
correct

 

[nycmathguy]

Q. 10

The horizontal asymptote is the line y = 0-> correct
The vertical asymptote is the line x =2
View attachment 487


12.
Vertical: x=- 3/2
Horizontal: y=-7/2 -> you did it correct

Q. 14
no horizontal asymptote=> correct
Vertical: x= -2

Q. 16
correct

Precalculus is getting interesting, to say the least. I thank you for taking the ride with me.