x=2 and x=-2
Vertical asymptotes concerns the bottom factors after simplification.
This cannot be simplified any further because x^2-4=(x-2)(x+2)
which means it has no common factors with the numerator (the top part of fraction)
So we will consider when both factors are 0.
x-2=0 when x=2
x+2=0 when x=-2
So the VA (Vertical asymptotes) are x=2 and x=-2
You must actually write x=2 and x=-2
Do not put just 2 and -2
So, the vertical asymptotes are
(b) x = -2/3
The vertical asymptote is the restriction on x.
The denominator cannot be equal to zero so that it the restriction.
Set the denominator equal to zero and solve for x to find the asymptote.
3x + 2 = 0
3x = -2
x = -2/3
1/2 is your vertical asymptote
17. 2x^4 + x^3 + x^2 + 4x + 3 + 8/(x -1)
18. x = 2, x = -2
19. The slant asymptote is y = x + 14Explanation:
17. See the first attachment for the synthetic division. The summary statement is the expression represented by the result.
18. The denominator is the difference of two squares, so is readily factored to ...
... (x -2)(x +2)
The zeros of this product are the locations of the vertical asymptotes of f(x). They are ...
... x = 2, x = -2.
19. Dividing the numerator by the denominator (using synthetic division, if you like) gives the result ...
... f(x) = x + 14 + 104/(x -8)
The linear expression y=x+14 defines the end behavior when x gets large. That is, it is the slant asymptote of the function. See the second attachment for a graph.
(There will be a horizontal asymptote only when the degrees of numerator and denominator are the same. In this case, they are not.)
There is a vertical asymptote for the rational function at x = −7. Set the denominator equal to 0 and solve for x.
x + 7 = 0 → x = −7
Vertical asymptotes happen at x=a if the function is undefined at x=a.
This is a rational functions, and rational functions are not defined when the denominator is zero, since you can't divide by zero.
In this case, the denominator is zero if
And thus the function has a vertical asymptote at x= -3/2
x = 3, x = -3
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non zero for these values then they are vertical asymptotes.
Solve : x² - 9 = 0 ⇒ x² = 9 ⇒ x = ± 3
The vertical asymptotes are x = - 3 and x = 3
The denominator 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.
2x - 3 = 0 ⇒ 2x = 3 ⇒ x =
Thus x = is the vertical asymptote
For the vertical asymptote test in a rational function. The denominator is first set to zero.
x^2 - 9 = 0
x^2 = 9
x = +-3
The values for x are then substituted to the numerator.
x=3; x-6(x+6) = 3-6(3+6) = -51
x=-3; x-6(x+6) = -3-6(-3+6)= -21
The zeros of the denominator do not make the numerator zero so the vertical asymptotes are x= +-3