Question

HELPP!!!
Identify the horizontal asymptote of f(x) = quantity 6 x minus 7 over quantity 11 x plus 8.

y = 0
y = 6 over 11
y = negative 7 over 8
No horizontal asymptote

Answer 1

\(\LARGE\color{black}{ f(x)= \frac{6x-7}{11x+8} }\)

Answer 2

Lets make an observation.
at first, you know that
\(\LARGE\color{black}{ 11x+8\neq0 }\)
and this value that x can't equal is the vertical asymptote.

Answer 3

Can you find the vertical asymptote?

Answer 4

idk how which is why i mainly need help with this

Answer 5

the denominator can't equal zero, you know this, right?

Answer 6

yes

Answer 7

because we can never divide by 0

Answer 8

right

Answer 9

So, this x-value, where the denominator is zero, is where the function doesn't have any value. It is undefined there.

Answer 10

Okay?

Answer 11

ok

Answer 12

lets find this value.
11x+8=0 (where does the function NOT exist)
subtracting 8 from both sides
11x=-8
dividing by 11 on both sides,
x=-8/11
this is the vertical asymptote.

Answer 13

good so far?

Answer 14

yes :)

Answer 15

I'll tell you hat does it mean to have a vertical asymptote.

Answer 16

is

Answer 17

okay :)

Answer 18

If you try to hit an x-value that is close to this x=-8/11
will approximate it to -0.73.
so, our asymptote is x=-0.73

if you try x=-0.72
you will get a value that is a little bit greater than zero.
So the closer you are getting to -0.73 (the -8/11) the smaller the denominator will be right?

Answer 19

This means that as x approaches -8/11, the y gets infinitely large. (keep in mind, it doesn't rich exactly -8/11, because there is no value there)

Answer 20

I'll give you time to process this.

Answer 21

ask any why questions if you don't get something in the middle of the text....

Answer 22

right i know that the closer we're getting to -8/11 the smaller the denominator will be

Answer 23

yes, so a very small positive decimal (like 0.00001) will make the f(x) 'S value infinitely large. this is when we choose an x value that is a tiny bit more than -8/11.

(by more I keep in mind that -4<-3 or -2<-1... you know what I mean? a larger negative number like -5, is SMALLER than a small negative number like -4)

Answer 24

So from the right, as x gets closer to -8/11, the y goes infinitely up and up.

Answer 25

yes

Answer 26

BUT, the top is negative. because the top is "6x-7" so when we plug in something close to -8/11 (or any negative value) for x, the numerator is negative.

So... For numbers that are a big greater than -8/11,

while we get a small positive decimal on the bottom, on the top though we get a fairly big negative number.

And


big negative number
----------------- = NEGATIVE INFINITY (ok ?)
small positive decimal

Answer 27

right

Answer 28

Okay.
We have observed that.
(This is actually a calculus-wise understanding of an asymptote)

Answer 29

LOL wooow

Answer 30

Now, lets go from the other side of the asymptote.
We have plugged in numbers that are on the right of the asymptote. And this gave us a picture like,