Question

Anyone getting A?
Please and thank ya! If the graph below is translated four units to the left and six units up, what would be the
rule of the new function?

Answer 1

i know the answer is A but can someone work it for me?

Answer 2

so it has a vertical asymptote at -9 and a horizontal asymptote at 3

Answer 3

how did u know that?

Answer 4

looking at the graph then counting from the original asymptote

Answer 5

so how owuld u do it?

Answer 6

BTW the answer isn't A

Answer 7

the back of the book said it was :-/

Answer 8

The only reason I saw it isn't A is because the asymptote occurs at -9 which means that -9 would not be in the domain of the function

Answer 9

the vertical asymptote that is

Answer 10

Remember you can't divide by 0, so if you put -9 into any of those functions which one would result in you dividing by 0

Answer 11

that is the one that will give you the vertical asymptote because remember an asymptote arises when a graph can get infinitely close to a number but never reach it

Answer 12

A would probably be the original function before you translated it

Answer 13

what do u thikink the anser is

Answer 14

The answer is D, the person who put the answers in your book I think made a mistake and didn't translate the graph

Answer 15

or they changed the question last minute

Answer 16

to include the translation

Answer 17

i got d first too thats why i was confuesd!

Answer 18

the reason why there is a horizontal asymptote is because

y = 1/(x-9) + 3

if you take the

limit as x-> infinity, you get y = 0 + 3, thus there is an asymptote at y = 3

Answer 19

or

y = 1/(x-9) + 3 can never equal to 3

I hate how they simplified it, im going to turn it into a fraction to better explain

y = (1 + 3(x-9))/(x-9) = ( 3x+ 1 + 3(-9) )/(x-9)
best way to spot a horizontal asymptote is to eliminate all numbers just leaving the highest degree variable in the equation so

3x/x = 3 thus there is an asymptote at 3, as the equation can never equal 3

if you get something like
1/x^2
or
1/x
etc...

the asymptote is equal to 0

this all related to taking the limit as x goes to infinity

Answer 20

but yeah the text book is wrong

Answer 21

hope that helps

Answer 22

if you get a function where x is larger on the numerator like

x^2/x
or even
x
there is no horizontal asymptote.

These aren't proofs btw so dont expect to get marks on the test for them but they are a quick way to check

Answer 23

ok thanks!

Answer 24

These videos explain it better than I, I recommend watching them and maybe taking a note or two they can really help

Horizontal asymptotes
https://www.youtube.com/watch?v=c-yK2hUnSB0&feature=relmfu

Vertical asymptotes
https://www.youtube.com/watch?v=_qEOZNPce60