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VikaD [51]
3 years ago
5

What are the vertical asymptotes of 10/X-1​

Mathematics
1 answer:
anastassius [24]3 years ago
5 0

Answer:

Enter a problem...

Algebra Examples

Popular Problems Algebra Find the Asymptotes f(x)=10/(x^2-1)

f

(

x

)

=

10

x

2

−

1

Find where the expression

10

x

2

−

1

is undefined.

x

=

−

1

,

x

=

1

Since

10

x

2

−

1

→

∞

as

x

→

−

1

from the left and

10

x

2

−

1

→

−

∞

as

x

→

−

1

from the right, then

x

=

−

1

is a vertical asymptote.

x

=

−

1

Since

10

x

2

−

1

→

−

∞

as

x

→

1

from the left and

10

x

2

−

1

→

∞

as

x

→

1

from the right, then

x

=

1

is a vertical asymptote.

x

=

1

List all of the vertical asymptotes:

x

=

−

1

,

1

Consider the rational function

R

(

x

)

=

a

x

n

b

x

m

where

n

is the degree of the numerator and

m

is the degree of the denominator.

1. If

n

<

m

, then the x-axis,

y

=

0

, is the horizontal asymptote.

2. If

n

=

m

, then the horizontal asymptote is the line

y

=

a

b

.

3. If

n

>

m

, then there is no horizontal asymptote (there is an oblique asymptote).

Find

n

and

m

.

n

=

0

m

=

2

Since

n

<

m

, the x-axis,

y

=

0

, is the horizontal asymptote.

y

=

0

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

This is the set of all asymptotes.

Vertical Asymptotes:

x

=

−

1

,

1

Horizontal Asymptotes:

y

=

0

No Oblique Asymptotes

image of graph

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DaniilM [7]

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3 years ago
Read 2 more answers
A cement walkway of uniform width has been built around an in-ground rectangular pool. the area of the walkway is 1344 square fe
Elza [17]

Answer:

6 feet.

Step-by-step explanation:

Dimensions of the Pool =80 feet long by 20 feet wide

Area of the walkway =1344 square feet.

If the width of the walkway=w

  • Length of the Larger Rectangle =80+2w
  • Width of the Larger Rectangle =20+2w

Area of the Walkway =Area of the Larger Rectangle - Area of the Pool

1344=(80+2w)(20+2w)-(80*20)\\1344=80*20+160w+40w+4w^2-80*20\\4w^2+200w=1344\\4w^2+200w-1344=0\\We factorize\\4(w^2+50w-336)=0\\4(w^2-6w+56w-336)=0\\w(w-6)+56(w-6)=0\\(w-6)(w+56)=0\\w-6=0$ or $ w+56=0\\w=6$ ft or $ w=-56$ \\Therefore, the width of the walkway , w=6 feet.

6 0
3 years ago
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