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svetlana [45]
3 years ago
11

50PTS! Describe the vertical asymptotes and holes for the graph of y = x-5/x^2 -1

Mathematics
2 answers:
xeze [42]3 years ago
8 0

Answer:

yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

Step-by-step explanation:

Diano4ka-milaya [45]3 years ago
3 0
Ok

first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0

then simplify
any factors you factored out is now a hole, remember them

to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve

so

y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0

no factors to simplify so no holes

set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1

x+1=0
x=-1

the horizontal assemtotes are x=1 and -1
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Answer:

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Step-by-step explanation:

<u>Complete Question:</u>

A car dealership has 24ft of dividers with which to enclose a rectangular play space in a corner of a customer lounge. The sides against the wall require no partition. Suppose the play space is x feet long.  

A) express the area of the play space as a function of x

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Solution:

A)

The rectangle has area of length * width

length is x

so width will be 24 - x

Hence, the area will be:

Area = x(24 - x)

Area = A(x) = 24x - x^2

B)

Domain is the set of x values that make the function defined.

The domain will be all values less than 24.

Because if you take 24, the area would be 0 [doesn't make sense]

if you take anything over 24, the area would be negative [not possible]

Hence, the domain is:

Domain = {x < 24}

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