Question

how do i find the vertical asymptote of the function f(x)=4/(x+5)

Answer 1

you first cross out like terms from top to bottom, then what ever value makes the bottom a zero is your VA

Answer 2

in this case....x =-5

Answer 3

so you just make x+5=0

Answer 4

then solve for x?

Answer 5

Try finding the lim as x goes to 0

Answer 6

that is correct, you want to know what vale of x can never be used

Answer 7

x=-5 is a nono in this equation, so never use it, draw a lint to mark its position and stay away from it, that your VA

Answer 8

line not lint, but if it is in macaroni art, use zitis

Answer 9

so to find the equation of the vertical asymptote of the function f(x)=4/x+5
what is the answer?

Answer 10

i am so confused?

Answer 11

lie down on my couch right here and lets talk about this confusion :) tell me, can the bottom of a fraction ever be a zero?

Answer 12

no

Answer 13

then the line on a graph that marks the spot where the bottom would be zero is indictaed by the vertical asymptote. at the value of x that is bad.

Answer 14

what value of x makes the bottom of your fraction there go bad?

Answer 15

anything negative?

Answer 16

negative numbers are not bad numbers they are just misunderstood :) there is on value in particular that is going to be offlimits:

x+5 = 0 when x =?

Answer 17

-5

Answer 18

correct! so that is your VA

Answer 19

x = -5 is the VA

Answer 20

ok i get it now,

Answer 21

there i graphed in it matlab shows you what is happening in case the visual aspect helps

Answer 22

you got that in crayon? :)

Answer 23

then how do i get horizontal asym to 5x^2-4/(x+1)

Answer 24

we divide the top by the bottom like noremal long division to get an exact answer, but its not a horizontal asymptote.... itll be a line that is slanted

Answer 25

look at the graph that should help you visualize it.

Answer 26

the right side there seems off, but that might be the zoom limit

Answer 27

ill try to fix it

Answer 28

if we simply do a shortcut and divide all the numbers by x, we get the top to be 5x

Answer 29

there you go now its a bit better

Answer 30

thats better, but the your asymptotes gone parabolic instead of linear :)

Answer 31

5x^2-4/(x+1)

the top and bottom differ by 1 degree, so the asymptote is linear

Answer 32

sorry i had the equation not copy and pasted wrong lol here is the correct result

Answer 33

yay!! i knew you could do it :)

Answer 34

man im just trying to help him out?

Answer 35

youre doing great....

Answer 36

im pretty sure when you learn these things it helps to look at a graph

Answer 37

it does, i am visual alot, the analysis part still gets my brain tied in knots

Answer 38

sorry i thought you were being sarcastic there

Answer 39

;) wel it was in humour, but nothing cruel or hateful. we all have errors to keep us humble i beleive, i know i have my share

Answer 40

well to be honest i did have the right equation i just got excited using matlab and didn bracket off the exponents properly