Mathematics, 27.07.2019 06:20 dommalb

All asymptotes. f(x)= x^2+1/x. show work.

Answers: 1

Show answers

Answers

Answer from: binks3350

1st problem: Find all asymptotes of $f(x)=x^2+\frac{1}{x}$ .

Horizontal: none

Vertical: x=0

Slant: none

2nd problem: Find all asymptotes of $f(x)=\frac{x^2+1}{x}$ .

Horizontal: none

Vertical: x=0

Slant: y=x

Step-by-step explanation:

I'm going to do two problems:

1st problem: Find all asymptotes of $f(x)=x^2+\frac{1}{x}$ .

2nd problem: Find all asymptotes of $f(x)=\frac{x^2+1}{x}$ .

Problem 1:

$f(x)=x^2+\frac{1}{x}$

I'm going to combine the terms; find a common denominator.

This means I will need to multiply first term by x/x:

$f(x)=\frac{x^3}{x}+\frac{1}{x}$

$f(x)=\frac{x^3+1}{x}$

Horizontal asymptotes will not exist if the degree of the numerator is larger than the degree of the denominator.

This is the case here because the degree of the numerator is 3 while the degree of the denominator is 1.

Slant asymptotes will only exist when the degree of the numerator is one more than the degree of the denominator. Since 3 is not one more than 1, then we don't have any slant aymptotes.

The vertical asymptotes will cause the denominator to be 0 while the numerator will be another number other than 0.

x is 0 when x=0.

We plug into top, 0^3+1=1.

x=0 gives us 1/0 so you definitely have a vertical asymptote at x=0.

Problem 2:

There are no horizontal asymptotes again because degree of top is more than degree of bottom.

Slant asymptote will exist when the degree of the top is 1 more than the degree of the bottom which is here.

The degree of the top is 2 while the degree of the bottom is 1.

Let's divide to find the slant aysmptote:

$\frac{x^2+1}{x}=\frac{x^2}{x}+\frac{1}{x}=x+\frac{1}{x}$

So the slant aysmptote will be the quotient part of this which is x.

So the slant asymptote is y=x.

The vertical asymptote will happen when denominator is 0 and the top is something that is not 0. (It should remain this way after simplifying as well).

So again we are dividing just by x.

So x is 0 when x=0.

Since 0^2+1=1 and not 0 then you have a vertical asymptote at x=0.

Answer from: Quest

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft

Answer from: Quest

maya’s expression will generate the wrong sum because she violated the commutative property.

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 17:30

What is not true about kl and mn? what is mn? (show your work)

Answers: 1

Answer

Mathematics, 21.06.2019 18:00

Me, prove a quadrilateral with vertices g(1,-1), h(5,1), i(4,3) and j(0,1) is a rectangle using the parallelogram method and a rectangle method.

Answers: 3

Answer

Mathematics, 21.06.2019 20:30

In an improper fraction, is the numerator equal to, less than, or greater than the denominator

Answers: 2

Answer

Mathematics, 21.06.2019 21:30

Tom can paint the fence in 12 hours, but if he works together with a friend they can finish the job in 8 hours. how long would it take for his friend to paint this fence alone?

Answers: 1

Answer

You know the right answer?

All asymptotes. f(x)= x^2+1/x. show work....

Questions