Question

PLEASE HELP
Identify the asymptotes, domain, and range of the function. Check all that apply.
f(x)=1/(x+4)+3

asymptotes: x = -4, y = 3
R:{y|y≠3}
asymptotes: x = 3, y = -4
D:{x|x≠−4}
R:{y|y≠−4}
D:{x|x≠3}

Answer 1

\(\displaystyle f(x)=\frac{1}{x+11}-8\)

(1) vertical shifts \(8\) units down,
(2) vertical asymptote at \(x=-11\), and
(3) horizontal asymptote at \(x=-8\).

Answer 2

Why? I assume you know (1)!

Reason (2)
\(x+11\ne0\), because otherwise you are dividing by zero.
So, \(\displaystyle \frac{1}{x+11}\ne 0\).

Note then, that if \(\displaystyle g(x)=\frac{1}{x+11}\), then you therefore have a horizontal asymptote at \(x=0\) (because \(g(x)\ne0\), however the closer you tend to zero, the bigger value you get, going into \(\pm\) infinities on both sides of \(y=0\).

Then, \(\displaystyle f(x)=\frac{1}{x+11}-8\) is same, but all of this reasoning is shifted to \(y=-8\).

Answer 3

and vertical asymptote is relevantly simple.
if not the concept, for the very least you know that \(x=-11\) gives you a vertical asymptote, because if \(x=-11\) then \(x+11=0\) which is not allowed.

Answer 4

(not allowed due to the fact that you are dividing by 0)

Answer 5

How am I supposed to find what y=?

Answer 6

how are you supposed to find the horizontal asymptote?

Answer 7

Yes?

Answer 8

I've never heard that term before

Answer 9

OK, fine .... :)

Answer 10

how about "vertical asymptote" ?

Answer 11

I've heard that one

Answer 12

The vertical asymptote would be -4?

Answer 13

We will plot a function
\(\displaystyle f(x)=\frac{1}{x-2}+5\)
https://www.desmos.com/calculator/ve1u34ikjs

Answer 14

Let's make a flashback.

Answer 15

What is smaller 1/100 or 1/9999 ? (And why?)

Answer 16

1/9999 because it's a smaller odd

Answer 17

WRONG !

Answer 18

1/9999 is smaller because you are dividing by a larger number.

Answer 19

that is the way to think about it.

Answer 20

Right? The larger number you divide by, the smaller the result.
and the smaller number you divide by, the bigger the result. AGREE ?

Answer 21

Yes, I agree. 1/100 would be a smaller decimal than 1/9999

Answer 22

Sorry, larger

Answer 23

I was almost despaired before you fixed that :)

Answer 24

Haha

Answer 25

OK, so here we come to the reason why division by 0 is invalid ...

Answer 26

1/0.1=10
1/0.05=20
1/0.009=1250
....

Answer 27

And this is making sense, right (?)
Because the smaller the number we divide by, the larger the result, right?

Answer 28

Note as well, that you can choose a number that is very very close to zero, and there are infinitely many (real) numbers that are going closer and closer to zero, right?

Answer 29

Yes

Answer 30

It's making sense

Answer 31

So, the closer x is to 0, the bigger the quotient 1/x is.

Answer 32

Well, for any number that is very close to zero you may think, I can come up with the number that is closer to zero.

(For the very least part, this is true, because if your number is \(x\), then \(y=x/2\) is closer to zero.)

Answer 33

So, this x that approaches 0, will lower than anything, and thus (intuitively) the quotient 1/x is going to be bigger than anything (i.e. infinity).

Answer 34

and the logic is the same, for negatives.
In other words, if you approach 0 from the negative side (e.g. -0.0004 and smaller values on that are tending to 0)

Answer 35

with an exception that, a very "small" negative decimal (that is smaller than anything) (such as -0.0000000400123) (will call it x) when 1/x, you get negative infinity.