Vertical Asymptotes
Find the vertical asymptote(s) of 2(x-6) / (x-6)(x+2)
x=-2
What is a Vertical Asymptote?
An imaginary line that a function approaches but never touches
The horizontal asymptote for x3+x+5/x2+4 is located....
None. There is not vertical asymptote
Find the HA and VA:
y= x2+3x+2 / x2+5x+6
HA: y= 1
VA : x = -3
Find the vertical asymptote(s) 2 / x(x-1)
x = 0 , x = 1
Sara said that the equation (x-2)(x+3) / (x +2)(x - 6) has one VA. Is she correct or incorrect? If she's correct why, if she's incorrect fix her answer and why.
Sara is incorrect. None of the factors in the numerator or denominator cancel out, so that means there will be 2 VA asymptotes.
The horizontal asymptote for x2+x+5/3x2+9 is located....
What is y=1/3
Find the HA and VA:
2x3-4x2+3x-12 / x2-16
HA: None
VA: x = -4
Find the vertical asymptote(s). (x+2)(x-4) / 2(x+2)(x-4)
None
Why do Vertical Asymptotes exist in rational functions?
They exist because a fraction's denominator can never be equal to 0
The horizontal asymptote for 3x6+x+5/2x9+4 is located....
What is y = 0
Find the HA and VA:
3x2-2x-8/x2+2x-8
HA: y=3
VA: x=-4
Find the vertical asymptote. y= 2x2-2x / (x -1)(x-2)
x = 2
At which value of the denominator is a VA created?
0
The horizontal asymptote for (x+3)(x-2)(x-4)/x(x+2) is located....
What is no Horizontal asymptote. The degree of the numerator is larger than the degree of the denominator.
Find the HA and VA:
2x3+6x2-2x-6 / x3+3x2-x-3
HA: y = 2
VA: None
Find the vertical asymptote. y= x3 -4x / (x-2)(x-4)
x = 4
What are the steps to find the vertical asymptote(s) of a rational function? (YOU MUST SAY ALL THE STEPS!)
1. Factor numerator and denominator.
2. Cancel out the like factors in the numerator and the denominator.
3. Whatever is left over in the denominator, set that = 0 and solve to find your VA.
The horizontal asymptote for (2x+4)(x-5)/(x3+2)(x-1) is located....
What is y=2
Find the HA and VA:
f(x)= (x3-16x) / -4x2+4x+24
HA: None
VA: x=3 and x=-2