Question

Determine the vertical and horizontal asymptote if one exists. graph. f(x) = 5x^2 - 5 / x^2 - 25.

Answer 1

i got H.A at 5. and v.a at 5 and -5

Answer 2

factor the top and bottom to see if any terms cancel; and whats leftover can be calculated for the va

Answer 3

ha = 5 yes

Answer 4

my graph is a parabola facing downward at the points of (1,0) and (-1,0)

Answer 5

5(x+1)(x-1)
-----------
(x+5)(x-5)

nothing cancels so yep; vas = 5 and -5

Answer 6

there is a wider view of this tho:

http://www.wolframalpha.com/input/?i=%285x^2+-+5%29+%2F+%28x^2+-+25%29

Answer 7

rainbow link maybe?

\begin{array}l\color{red}{\normalsize\text{h}}\color{orange}{\normalsize\text{t}}\color{#9c9a2e}{\normalsize\text{t}}\color{green}{\normalsize\text{p}}\color{blue}{\normalsize\text{:}}\color{purple}{\normalsize\text{/}}\color{purple}{\normalsize\text{/}}\color{red}{\normalsize\text{w}}\color{orange}{\normalsize\text{w}}\color{#9c9a2e}{\normalsize\text{w}}\color{green}{\normalsize\text{.}}\color{blue}{\normalsize\text{w}}\color{purple}{\normalsize\text{o}}\color{purple}{\normalsize\text{l}}\color{red}{\normalsize\text{f}}\color{orange}{\normalsize\text{r}}\color{#9c9a2e}{\normalsize\text{a}}\color{green}{\normalsize\text{m}}\color{blue}{\normalsize\text{a}}\color{purple}{\normalsize\text{l}}\color{purple}{\normalsize\text{p}}\color{red}{\normalsize\text{h}}\color{orange}{\normalsize\text{a}}\color{#9c9a2e}{\normalsize\text{ }}\\\color{green}{\normalsize\text{c}}\color{blue}{\normalsize\text{o}}\color{purple}{\normalsize\text{m}}\color{purple}{\normalsize\text{/}}\color{red}{\normalsize\text{i}}\color{orange}{\normalsize\text{n}}\color{#9c9a2e}{\normalsize\text{p}}\color{green}{\normalsize\text{u}}\color{blue}{\normalsize\text{t}}\color{purple}{\normalsize\text{/}}\color{purple}{\normalsize\text{?}}\color{red}{\normalsize\text{i}}\color{orange}{\normalsize\text{=}}\color{#9c9a2e}{\normalsize\text{%}}\color{green}{\normalsize\text{2}}\color{blue}{\normalsize\text{8}}\color{purple}{\normalsize\text{5}}\color{purple}{\normalsize\text{x}}\color{red}{\normalsize\text{^}}\color{orange}{\normalsize\text{2}}\color{#9c9a2e}{\normalsize\text{+}}\color{green}{\normalsize\text{-}}\color{blue}{\normalsize\text{+}}\color{purple}{\normalsize\text{5}}\color{purple}{\normalsize\text{%}}\color{red}{\normalsize\text{2}}\color{orange}{\normalsize\text{9}}\color{#9c9a2e}{\normalsize\text{+}}\color{green}{\normalsize\text{%}}\color{blue}{\normalsize\text{2}}\color{purple}{\normalsize\text{F}}\color{purple}{\normalsize\text{+}}\color{red}{\normalsize\text{%}}\color{orange}{\normalsize\text{2}}\color{#9c9a2e}{\normalsize\text{8}}\color{green}{\normalsize\text{x}}\color{blue}{\normalsize\text{^}}\color{purple}{\normalsize\text{2}}\color{purple}{\normalsize\text{+}}\color{red}{\normalsize\text{-}}\color{orange}{\normalsize\text{+}}\color{#9c9a2e}{\normalsize\text{2}}\color{green}{\normalsize\text{5}}\color{blue}{\normalsize\text{%}}\color{purple}{\normalsize\text{2}}\color{purple}{\normalsize\text{9}}\color{red}{\normalsize\text{}}\end{array}

Answer 8

:) that's pretty

Answer 9

lol ... it doesnt function when I contort it like that :)

Answer 10

how do i know weather or not there are lines around my parabola or not?

Answer 11

a vertical asymptote indicates that there is something going on further down the line. a va is in the middle of a graph, not towards the ends usually

Answer 12

as long as the domain of the function is beyond the va; its gonna have extra parts :)

Answer 13

so am i supposed to have something more than an upside down parabola on my graph? other than the ha and va?

Answer 14

yep; look at the area on the other side of the vas where the x values keep going ...

Answer 15

\begin{array}c
&&&.&&.\\
&&&.&&.\\
&&&.&&.\\
.&.&.&&&&.&.&.\\
&&&&..&\\
&&&.&&.\\
&&&.&&.\\
&&&.&&.\\
\end{array}

Answer 16

wow, how did you do that?

Answer 17

very carefully :)

I just coded the editor to line up an array ..

Answer 18

:) that's awesome

Answer 19

\begin{array}c
&&&.&&.\\
&&&.&&.\\
&&&.&&.\\
.&.&.&&&&.&.&.\\
&&&&..&\\
&&&.&&.\\
&&&.&&.\\
&&&.&&.\\
\end{array*}

Answer 20

impressive

Answer 21

if I had time i could prolly work that into a neat little graphing option for th esite :)

Answer 22

:)

Answer 23

^
|
<___|

Answer 24

:) the array option helps to keep things aligned

Answer 25

lol, i don't know what i'm doing. is this what goes on the left side of the vertical asymptote?

Answer 26

and it's mirror on the other side?

Answer 27

yes, something akin to that; and the mirror of its on the right

Answer 28

okay. thank you sensei. :)

Answer 29

youre welcome :) good luck!!

Answer 30

:)