Question

GRAPHING QUESTION PLS HALP

Answer 1

Do you know how to calculate VA?

Answer 2

Set denominator equal to zero so VA @ x=-2

Answer 3

no i do not

Answer 4

Can you show me how to do this so I can do the other ones myself

Answer 5

Sure
\[x+2=0\]

x=-2

Vertical asymptote @ x=-2

Why? Because if x=-2 then you would be dividing by zero which is impossible, so the graph will "not be defined" at x=-2 so there is an asymptote.

Answer 6

https://2.bp.blogspot.com/-9uxn3U5jEz8/VC-TK4qZudI/AAAAAAAAAGc/zt-dgnMHBfw/s1600/Screen+Shot+2014-10-03+at+11.26.10+PM.png

Answer 7

ok

Answer 8

so whats next

Answer 9

Check out that image for horizontal asymptotes. The exponets are key for HA's

If the exponent in the numerator is bigger then the horizontal asymptote is y=0, or in other words, the x-axis. If the exponent in the denominator is bigger then there is no horizontal asymptote. If both the exponets are the same then the HA is the coffecients in front of the x's for example...\[y=\frac{ 13x^3 }{ 21x^3 }\]
For this equation, because both of our x's exponents are equal (x to the third) our horizontal asymptote would be 13 over 21 or

\[\frac{ 13 }{ 21 }\]

Answer 10

ohhhhhh

Answer 11

I mostly get it now

Answer 12

In this case the number in the numerator is bugger so the x-axis would be the HA

Answer 13

so how would I graph that

Answer 14

i get the concept, just dont know how to graph it

Answer 15

then just plug in a couple random numbers into the equation and plot them

Answer 16

for example when you plug in zero for x what do you get

Answer 17

like any numbers

Answer 18

2.5

Answer 19

yep

Answer 20

So one point would be (0, 2.5)

Answer 21

do i need a horizontal asymptote

Answer 22

id write HA on top of the x-axis so your teacher / professor knows

Answer 23

@legomyego180 this is what my graph looks like though

Answer 24

there is a horizontal asymptote, its just on the x-axis so its hard to see.

This is how the graph should look:

Answer 25

that graph is incorrect

Answer 26

When you graph make sure you enter it just like this:

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

Answer 27

idk why but the system is not letting me

Answer 28

try wolframalpha.com or mathway.com

Answer 29

is there a vertical asymptote

Answer 30

yea. If you can plug and number into x to make the denominator zero there will be a vertical asymptote. The reason we have vertical asymptotes is to keep zeros out of the denominator.

Functions with Vertical asymptotes:

\[y=\frac{ x+43 }{ x-1 }\]
Vertical Asymptote @ x=1

\[y=\frac{ x-43232342+19-42 }{ x+3 }\]
Vertical asymptote at x=-3

Funtcions without VA's:
\[\frac{ 14+3x }{ 2 }\]
You cant get an x in the denominator here, its just two, its not gona change

\[\frac{ 3+14x^2 }{ x^2+2 }\]

Cant get a vertical asymptote here because x^2 is always positive and even if you plug in 0 for x you will still get 2...in other words...
\[x^2+2\neq0\]

Answer 31

If you plug a number in for x and it makes the denominator 0 you have a vertical asymptote running straight up and down at that x-value. It's a very important concept to understand - you talk about it a lot for the rest of math.