Question

what is an asymptote?

Answer 1

its an imaganiry line that a curve never quite touches but come close to

Answer 2

its like a boundary

Answer 3

so if a vertical asymptote was x=2 then no points on the graph would corss it at any time

Answer 4

make sense?

Answer 5

an asymptote is a horizontal or vertical line that is approached by a graph without touching it.

Answer 6

Asymptotes need not be lines. An asymptote can be any function. For instance there are quadratic asymptotes.

Answer 7

and aproaching an infinite x or/and y value

Answer 8

the problem is identify the asymptotes, domain and range of the function g(x) = (1/x+7) +3

Answer 9

Asymptote could also be slant as well

Answer 10

an asymptote of a curve is always a line.

Answer 11

Technically, a graph can cross a horizontal asymptote (it's rare, but it can happen), but it can never touch a vertical asymptote.


The more accurate definition involves the use of limits (which is a calculus term and concept), so this definition works best (even though it has its flaws).

Answer 12

A function has an asymptote if the limiting behaviour of that function is that asymptote. An asymptote can be any other function.

Answer 13

you can use limits in getting the asymptote of the curve, but that's only recommended for more complex functions like piecewise functions, for example.

Answer 14

Guys I think we're too wrapped in the definition there's actually a problem to be solved: the problem is identify the asymptotes, domain and range of the function g(x) = (1/x+7) +3

Answer 15

To find any x - asymptotes you take the function in the denominator and set it equal to 0
x + 7 = 0 therefore x = -7

Answer 16

domain: all real numbers except -7

Answer 17

For the y asymptote it is the constant 3 so y = 3

Answer 18

"soxhletapparatus: an asymptote of a curve is always a line"

This is incorrect:
Example (quadratic asymptote):
\[f(x)=\frac{(x+3)^3}{x-3}\]

Answer 19

The domain is all real numbers except -7, of course because we have found out that the function is undefined at that point.

Answer 20

Soxhletappartauts will you go to the chat for a sec?

Answer 21

The range will be all real numbers

Answer 22

Vertical asymptote: \[\large x=-7\] (let the denominator x+7 equal zero and solve for x)

Horizontal asymptote: \[\large y=3\] (combine the terms and divide the leading coefficients)

Answer 23

combine which terms?

Answer 24

Combine \[\frac{1}{x+7}\] and 3 to get

\[\frac{1}{x+7}+3=\frac{1}{x+7}+\frac{3(x+7)}{x+7}=\frac{1}{x+7}+\frac{3x+21}{x+7}=\frac{1+3x+21}{x+7}=\frac{3x+22}{x+7}\]


So after combining the terms, we get \[\frac{1}{x+7}+3=\frac{3x+22}{x+7}\]


Now divide the leading coefficient in the numerator (3) by the leading coefficient in the denominator (1) to get 3/1 = 3

Answer 25

im sorry. im confused about that long equation up there

Answer 26

Basically, I added \[\frac{1}{x+7}\] to 3 and I got \[\frac{3x+22}{x+7}\]

That long equation shows I did that.

Answer 27

so you added 1/x+7 to 3 and multiplied the top and bottom by x+7?

Answer 28

\[\large \frac{1}{x+7}+3\]



\[\large \frac{1}{x+7}+\frac{3(x+7)}{x+7}\]



\[\large \frac{1}{x+7}+\frac{3x+21}{x+7}\]



\[\large \frac{1+3x+21}{x+7}\]



\[\large \frac{3x+22}{x+7}\]



So \[\large \frac{1}{x+7}+3=\frac{3x+22}{x+7}\]

Answer 29

Yes, that's what I did (shown in step 2). Hopefully that last thing I posted is a bit clearer.

Answer 30

ok. so all of that would equal 3x+21/x+7 right?

Answer 31

\[\large \frac{3x+22}{x+7}\] but you have the right idea

Answer 32

So this means that the function g(x) is now \[\large g(x)=\frac{3x+22}{x+7}\]

Answer 33

oh! gotcha. ok so now i have that. now what?

Answer 34

Now just divide the leading coefficients. The leading coefficients are the first coefficients. In the numerator, it's a 3. In the denominator, it's a 1 (since x=1x)


So \[\frac{3}{1}=3\]

This means that the horizontal asymptote is \[\large y=3\]

Answer 35

sweet. can you help me with finding the domain and range too?

Answer 36

sure, the domain is the set of all allowable inputs. Keep in mind that you CANNOT divide by zero


So this means that the denominator x+7 CANNOT be zero. If x+7 was zero then...


x+7=0
x+7-7=0-7
x=-7

So if x=-7, then x+7 is zero. So we have to exclude that value from the domain (since it causes a division by zero error)

This then means that the domain is \[\large \{x|x\in\mathbb{R},x\neq-7\}\]

This says: the domain is the set of all real numbers BUT x CANNOT equal -7

In interval notation, the domain is \[\large (-\infty,-7)\cup(-7,\infty)\]

Answer 37

The range is the set of possible outputs. The easiest way to find the range is to look at the horizontal asymptote. Often, the graph will not cross the horizontal asymptote (it can sometimes, but that's another story)


So because the horizontal asymptote is y = 3, this means that 3 is likely to be excluded from the range. Indeed, it turns out that y=3 is excluded from the range when we look at the graph of g(x)


So the range is \[\large \{y|y\in\mathbb{R},y\neq3\}\]


In interval notation, the range is \[\large (-\infty,3)\cup(3,\infty)\]

Answer 38

thank you so much!!!!