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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- amistre64

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

- amistre64

in this case....x =-5

- anonymous

so you just make x+5=0

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

then solve for x?

- anonymous

Try finding the lim as x goes to 0

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

i am so confused?

- amistre64

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?

- anonymous

no

- amistre64

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.

- amistre64

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

- anonymous

anything negative?

- amistre64

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 =?

- anonymous

-5

- amistre64

correct! so that is your VA

- amistre64

x = -5 is the VA

- anonymous

ok i get it now,

- anonymous

##### 1 Attachment

- anonymous

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

- amistre64

you got that in crayon? :)

- anonymous

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

- amistre64

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

- anonymous

##### 1 Attachment

- anonymous

look at the graph that should help you visualize it.

- amistre64

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

- anonymous

ill try to fix it

- amistre64

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

- anonymous

##### 1 Attachment

- anonymous

there you go now its a bit better

- amistre64

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

- amistre64

5x^2-4/(x+1)
the top and bottom differ by 1 degree, so the asymptote is linear

- anonymous

##### 1 Attachment

- anonymous

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

- amistre64

yay!! i knew you could do it :)

- anonymous

man im just trying to help him out?

- amistre64

youre doing great....

- anonymous

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

- amistre64

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

- anonymous

sorry i thought you were being sarcastic there

- amistre64

;) 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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.