Where are the asymptotesame for the following function located? f (x)=7/x^2-2x-24

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Where are the asymptotesame for the following function located? f (x)=7/x^2-2x-24

1) given function:

7

x^2 - 2x - 24

2) Vertical asymptotes are at the values ox x for which the function is not defined. These are the roots of the denominator polynomial.

You can factor the denominator to find those roots:

x^2 - 2x - 24 = 0 = (x - 6)(x + 4) => x = 6 and x = - 4

Therefore, the two vertical asymptotes are x = 6 and x = - 4

3) The function also has one horizontal asymptote:

Lim f(x) when x → +/- ∞ is 0, so y = 0 is also an asymptote.

x=-4 and x=6

Step-by-step explanation:

Factor the denominator of the equation and then set them equal to zero and you will get these answers.

Given:

f(x)=14/(x-5)(x+1)

The asymptotes for the following function is located in a. x=-1 and x=5.

x = -1 and x = 5 is the vetical asymptotes.

the horizontal aymptotes is located at y = 0.

there is no oblique asymptotes.

Pls. see attachment for addtl explanation.

[tex]Where are the asymptotes for the following function located? f(x)=14/(x-5)(x+1) a. x=-1 and x=5 b.[/tex]

F(x) = 14/(x - 5)(x + 1)

The vertical asymptotes are determined by the denominator.

Set each factor of the denominator equal to zero and solve.

x - 5 = 0 x + 1 =0

add 5 to both sides subtract 1 from both sides

x = 5 x = -1

LETTER A

Hence, the asymptotes of f(x) located at x=6 and x=-4.

Step-by-step explanation:

Since we have given that

[tex]f(x)= \frac{7}{x^2-2x-24}[/tex]

We need to find the asymptotes for the above function:

Asymptotes occur when denominator becomes zero.

[tex]x^2-2x-24=0\\\\x^2-6x+4x-24=0\\\\x(x-6)+4(x-6)=0\\\\(x-6)(x+4)=0\\\\x=6,-4[/tex]

Hence, the asymptotes of f(x) located at x=6 and x=-4.

Vertical aymptotes at x= -4 and x=6

horizontal asymptote at y=0

Step-by-step explanation:

[tex]f(x)= \frac{-7}{x^2-2x-24}[/tex]

To find out vertical asymptote we take the denominator =0 and solve for x

x^2-2x-24=0

sum = -2 and product = -24

4 * (-6) = -24

4 - 6 = -2

(x+4)(x-6)=0

Now we set each factor =0 and solve for x

x+ 4=0 and x-6=0

x= -4 and x=6

Vertical aymptotes at x= -4 and x=6

Now we find horizontal asymptote

The degree of numerator is 0

degree of denominator is 2

Degree of numerator is less than the degree of denominator then horizontal asymotote is y=0

0 < 2 so horizontal asymptote at y=0

Factor the demoniator:-

x^2 -2x - 24 = (x - 6)(x + 4)

the asymptotes occurs when denominator = 0

so here they are the vertical lines x = 6 and x = -4

2 vertical asymptotes occurring at x = 5 and x = -1

Step-by-step explanation:

given

[tex]f(x) = \frac{14}{(x-5)(x+1)}[/tex]

recall that asymptotic occur at the locations that will make the equation undefined. In this case, the asymptote will occur at x-locations which will cause the denominator to become zero (and hence undefined)

Equating the denominator to zero,

(x-5)(x+1) = 0

(x-5) =0

x = 5 (first asymptote)

or (x+1) = 0

x = -1 (2nd asymptote)

Option a is correct.

x=-1 and x=5

Step-by-step explanation:

To find the asymptotes for the rational function:

[tex]f(x) = \frac{14}{(x-5)(x+1)}[/tex]

The vertical asymptotes for this function is to set the denominator equals to 0.

The horizontal asymptotes for this given fucntion is 0.

Denominator of the given function is: [tex](x-5)(x+1)[/tex]

By definition of asymptotes;

[tex](x-5)(x+1) =0[/tex]

By zero product property:

[tex]x-5 = 0[/tex] and [tex]x+1 = 0[/tex]

⇒x = 5 and x= -1

Therefore, the asymptotes for the following function located are: x = -1 and x =5

Answer choice A.

x = –1 and x = 5