Question

Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

\[fx=\frac{ -2 }{ x+3} \]

Answer 2

I think y=0 and x=-3 but i dont see how that could be right....

Answer 3

What do you want to do with that?

Answer 4

Graph and find out the asymptotes

Answer 5

yes y=0 and x=-3 are the asymptotes

Answer 6

To plot this curve you should find some points

Answer 7

when x is a very negative number (say -1003) then y= -2/(-1003+3)= -2/-1000 = +1/500
or 0.002
so y is close to 0 but slight positive.

as x gets closer to 0, say x= -6 then y = -2(-6+3) = -2/-3 = 1.5

(-6, 3/2) is on the curve.

at x = -3.1 (not quite -3 ) then y= -2/(-3.1+3) = -2/-0.1= +20
(-3.1,20) is on the curve. the curve is shooting up approaching x= -3

Answer 8

What do you mean? Can you ask the question a different way?

Answer 9

like the inverse varation(k). xy=k and y=k/x, usualy i find k and plug in random 3s for y. Would that not work?

Answer 10

*plug in random #'s

Answer 11

it is the same idea, but you use
\[ y = \frac{-2}{x+3} \]
you can think of that as k is -2. but to find y you don't use "x", you use x+3

so to find y when x =0, you do : y= -2/(0+3) and simplify (to get -2/3)

Answer 12

But I would use the fact that the asymptotes are x=-3 and y=0 to help pick numbers

if you have an asymptote at x=-3 you should expect big # 's for y near x=-3
so try one number near x=-3 (example -2.5)

Answer 13

so x=1 y= -2/(1+3)?
y=-2/4
y=-1/2
Is this the basic way 2 do it?

Answer 14

yes , that is how to do it. you get the point (1, -0.5) to plot

Answer 15

Thank you so much!!

Answer 16

you will get 2 curves: one for when x is less than -3, and a separate "branch" for x>-3

Answer 17

ok

Answer 18

@satellite73 ?

Answer 19

is this right?