Vertical Asymptotes

[nycmathguy]

Section 2.6
10, 12, 14, 16

Question 10

Set denominator to 0 and solve for x.

(x - 2)^3 = 0

Cube root both sides.

cuberoot{(x - 2)^3} = cuberoot{0}

x - 2 = 0

x = 2

Question 12

Set denominator to 0 and solve for x.

3 + 2x = 0

2x = -3

x = -3/2

Question 14

Set denominator to 0 and solve for x.

x + 2 = 0

x = -2

Question 16

For this question, the quadratic formula is needed.

See attachment.

 

[MathLover1]

Question 10
Vertical: x=2-> you did this part
Horizontal: y=0

Question 12
Vertical: x= - 3/2-> you did this part
Horizontal: y=- 7/2

Question 14
Vertical: x=-2
Horizontal:
Do polynomial long division 4x^2/(x+2)=4x-8+16/(x+2)
y=4x-8 (slant asymptote)
A horizontal asymptote is a special case of a slant asymptote.

Question 16

as you can see from the graph, there is no vertical asymptote (function is continuous)


Horizontal:
f(x)=(-4x^2+1)/(x^2+x+3) horizontal is quotient of -4x^2/x^2=-4

=>horizontal asymptote is y=-4

 

[nycmathguy]

Question 10
Vertical: x=2-> you did this part
Horizontal: y=0

Question 12
Vertical: x= - 3/2-> you did this part
Horizontal: y=- 7/2

Question 14
Vertical: x=-2
Horizontal:
Do polynomial long division 4x^2/(x+2)=4x-8+16/(x+2)
y=4x-8 (slant asymptote)
A horizontal asymptote is a special case of a slant asymptote.

Question 16
View attachment 502

as you can see from the graph, there is no vertical asymptote (function is continuous)


Horizontal:
f(x)=(-4x^2+1)/(x^2+x+3) horizontal is quotient of -4x^2/x^2=-4

=>horizontal asymptote is y=-4

Why is my work for 16 wrong?

 

[MathLover1]

what you did in 16 is finding roots, and you needed too find asymptotes

 

[nycmathguy]

what you did in 16 is finding roots, and you needed too find asymptotes

This is what happens when trying to solve math problems with little to no sleep.