Enter the equations of the asymptotes for the function f(x)

f(x) = 3/ (x -7)+ 2

Vertical asymptote:

Horizontal asymptote:

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f(x) = 3/ (x -7)+ 2

Vertical asymptote:

Horizontal asymptote:

Vertical asymptote x= 1

Horizontal asymptote y= -1

Step-by-step explanation:

x = 7 and y = 2

Step-by-step explanation:

the denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote

solve x - 7 = 0 ⇒ x = 7 is the asymptote

horizontal asymptotes occur as

[tex]lim( x → ± ∞), f(x) → c ( where c is a constant )

divide terms on numerator/ denominator by x

f(x) = (3/x/x/x -7/x ) + 2

as x → ± ∞, f(x) → 0 / 1 - 0 + 2 = 2

y = 2 is the asymptote

V.A.: x = -4, H.A: y = -6

Step-by-step explanation:

Vertical Asymptote is the restriction on "x". Since the denominator cannot be equal to zero, then x + 4 ≠ 0 ⇒ x ≠ -4. So, the vertical asymptote is at x = 4.

Horizontal Asymptote is the restriction on "y". This is determined by the degree of the numerator compared to the degree of the denominator. In the given problem, the degree of the numerator is less than the degree of the denominator so the horizontal asymptote is y = 0. However, there is also a negative 6 after the rational expression so the horizontal asymptote is: y = 0 - 6 ⇒ y = -6

A)x=2pi/3 and x=-2pi/3

Step-by-step explanation:

The function [tex]y=\frac{5}{3}tan(\frac{3}{4}x)[/tex] has vertical asymptotes in the values where the tan(a) has vertical asymptotes.

we know that tan(a) has vertical asymptotes in [tex]a=\frac{\pi }{2}[/tex] and [tex]a=\frac{-\pi }{2}[/tex], if we made [tex]a=\frac{3x}{4}[/tex] and solve for x, we get:

for [tex]a=\frac{\pi }{2}[/tex]

[tex]\frac{\pi }{2} =\frac{3x}{4}\\x = \frac{2\pi }{3}[/tex]

for [tex]a=\frac{-\pi }{2}[/tex]

[tex]\frac{-\pi }{2} =\frac{3x}{4}\\x = \frac{-2\pi }{3}[/tex]

Finally, the function [tex]y=\frac{5}{3}tan(\frac{3}{4}x)[/tex] has vertical asymptotes in the values x=2pi/3 and x=-2pi/3

Vertical asymptote is x = 7

Horizontal asymptote is y = 2

Step-by-step explanation:

The vertical asymptote is the restriction on the domain (x-value). Since the denominator cannot be zero ⇒ x - 7 ≠ 0 ⇒ x ≠ 7 so the vertical asymptote is at x = 7.

The horizontal asymptote (H.A.) is the restriction on the range (y-value). There are three rules that determine the horizontal value which compare the degree of the numerator (n) with the degree of the denominator (m):

If n > m , then there is no H.A. (use long division to find the slant asymptote) If n = m , then the H.A. is the coefficient of n ÷ coefficient of mIf n < m, then the H.A. is y = 0

In the given problem, n = 0 and m = 1 ⇒ n < m ⇒ H.A. is y = 0

Since there is a vertical shift of +2 units, the H.A. is y = 0 + 2 ⇒ y = 2

No Vertical or Horizontal Asymptotes are found. See attachment:

[tex]Enter the equations of the asymptotes for the function f(x) . f(x)=4x−8−2 vertical asymptote: ?[/tex]

we are given

[tex]f(x)=(0.75)^x+2[/tex]

For finding asymptote , we can find limit

[tex]\lim_{x \to \infty} f(x)= \lim_{x \to \infty}((0.75)^x+2)[/tex]

[tex]\lim_{x \to \infty} f(x)= \lim_{x \to \infty} (0.75)^x+\lim_{x \to \infty} 2)[/tex]

now, we can solve it

[tex]\lim_{x \to \infty} f(x)= 0+2[/tex]

[tex]\lim_{x \to \infty} f(x)= 2[/tex]

so, horizontal asymptote is

[tex]y= 2[/tex].............Answer

we are given

[tex]f(x)=(0.75)^x+2[/tex]

For finding asymptote , we can find limit

[tex]\lim_{x \to \infty} f(x)= \lim_{x \to \infty}((0.75)^x+2)[/tex]

[tex]\lim_{x \to \infty} f(x)= \lim_{x \to \infty} (0.75)^x+\lim_{x \to \infty} 2)[/tex]

now, we can solve it

[tex]\lim_{x \to \infty} f(x)= 0+2[/tex]

[tex]\lim_{x \to \infty} f(x)= 2[/tex]

so, horizontal asymptote is

[tex]y= 2[/tex].............Answer

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x = 7, y = 2

Step-by-step explanation:

I assume it is 3/(x-7) + 2.

When x = 7, there is an asymptote because it is undefined.

When y = 2, there is also one, because 3/(x-7) is never 0.

These are the only ones.