Question

@DanJS

Answer 1

The vertical asmtotes are where the function will be undefined

Answer 2

\[\frac{ a }{ 0 } = undefined\]

Answer 3

so x/0=undefined?

Answer 4

f(x) will be undefined when

\[x^2 + 3=0~~~or~~~x^2+2 = 0\]

but is that possible?

Answer 5

right, dividing by a zero is not defined

Answer 6

wait no it isnt possible bc no rsult can make x^2+3=0

Answer 7

right, graph the thing too and see if you spot anywhere where the graph blows up to infinity or -infinity

Answer 8

so saying that is undefined then the answer will be that there is no vertical asymptotes?

Answer 9

Vertical asymptotes occur where the graph is undefined. Here there are no values of x that will make f(x) undefined.

Answer 10

If you had something like
\[f(x) = \frac{ 4 }{ x^2-4 }\]

then you will have vertical asmptotes at x=2 and x=-2

Answer 11

oh i see bc no matter what f(x)=1?

Answer 12

f(x) = 1 is a horizontal line at y=1

Answer 13

for the example i typed in above..
the bottom factors to
(x+1)(x-2) so if x=2 or x=-2 then the denominator will be zero

a zero denominator is undefined

Answer 14

(x+2)(x-2)

Answer 15

oooo ok that makes since

Answer 16

Horizontal asmytotes are a number that the graph approaches as you take very large values of x.
As x goes off to infinity, the graph approaches a value

Answer 17

Like
f(x) = 1/x

as x becomes very large. the fraction becomes very small and gets ever closer to zero.

F(x) = 1/x has a horizontal asmytote at y=0

Answer 18

and x=0 is a vertical asmytote

Answer 19

oh yes bc 3/4 is bigger than 1/25

Answer 20

so x=0 will always be the vertical asmtope?

Answer 21

\[f(x) = \frac{ 1 }{ x }\]

let x become larger and larger, f(x) = y becomes smaller and smaller,

For x goes to infinity, you can say that y goes to zero. y=0 is a horizontal asmytote