Question

Help With a Vertical asymptotes problem

Answer 1

23

Answer 2

\[\frac{x(x-7)}{x^3-49x}\] i get +-7

Answer 3

don't forget the 0 as well

Answer 4

hmmm wait a sec...

Answer 5

this is a test example and he put the answer as x=-7 wouldnt it be more then that?

Answer 6

but when i do the math for it i get +-7

Answer 7

the denominator's zeros will be the vertical asymptotes values, yes,
SO LONG they do not make the numerator zero

Answer 8

so +7 would make the numerator 0, so that's out

Answer 9

ohh
thats makes sense now

Answer 10

and zero is another btw... BUT 0, makes the numerator zero too, so that's out as well

Answer 11

mind explain another one to me? :P

Answer 12

ok

Answer 13

\[\frac{x^2-7}{49x-x^4}\] find the horizontal asymptote

Answer 14

the answer is 0 but not sure how to do

Answer 15

or setup

Answer 16

hmm ? what do you mean?

Answer 17

my answer for the worksheet says y=0 but i dont know how he got that

Answer 18

the rule is, is the denominator's degree is greater than the numerator's, then the HORIZONTAL asymptote is at y=0, or the x-axis notice \(\bf \cfrac{x^{\color{blue}{ 2}}-7}{49x-x^{\color{blue}{ 4}}}\)

Answer 19

so if i had a problem like \[\frac{x^2}{x^3}\] y=0 ?

Answer 20

yeap

Answer 21

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XjIveF4zIiwiY29sb3IiOiIjRTYxNTE1In0seyJ0eXBlIjoxMDAwfV0- <--- notice the horizontal asymptote

Answer 22

mind explain one more of the same problems then im done :P

Answer 23

ok

Answer 24

\[\frac{x^2+1x-8}{x-8}\] horizontal asymptote

Answer 25

how would i set this up with the degrees different ?

Answer 26

if the degree of the numerator's is greater, then there's no horizontal asymptote, \(\bf \cfrac{x^{\color{blue}{ 2}}+1x-8}{x^{\color{blue}{ 1}}-8}\)

Answer 27

he put the answer as y = x+9 which i dont get

Answer 28

well, yes, but that's an OBLIQUE or SLANTED asymptote, not a horizontal

rule is, if the numerator's degree is 1 more than the denominator's, then it has an oblique asymptote

the oblique asymptote has an equation from the quotient of the division of both
the numerator by the denominator \(\bf \cfrac{x^{\color{blue}{ 2}}+1x-8}{x^{\color{blue}{ 1}}-8}\qquad \qquad {\color{blue}{ 2}}={\color{blue}{ 1}}+1\\ \quad \\
\textit{oblique asymptote line equation will be } x^2+1x-8 \div x-8\)

Answer 29

ohh
i seee thank you so much for your help

Answer 30

yw