# Thread: Get everything from the graph

1. ## Get everything from the graph

$\displaystyle \mathfrak{hello}$
i'm trying to get all the possible informations about a function from its graph and i mean by informations;limits,asymptotes..
i hope someone would help me extract those informations from this example :
Let : $\displaystyle f : \mathbb{R}\to \mathbb{R}\setminus \left \{ -1 \right \}$
and here's its graph :

2. Hello, Raoh!

I'm trying to get all the possible informations about a function from its graph
and i mean by informations: limits, asymptotes . . .
It appears to have a vertical asymptote: $\displaystyle x = -1$
. . and a horizontal asymptote: $\displaystyle y = 1$

It also seems to be a hyperbola.

The hyperbola: .$\displaystyle y \:=\:\frac{a}{x}$ .has center (0,0) and appears in Quadrants 1 and 3.
The hyperbola: .$\displaystyle y \:=\:-\frac{a}{x}$ .has center (0,0) and appears in Quadrants 2 and 4.

Our hyperbola has center (-1, 1) and seems to contain: (3,0), (1,-1), (0,-3).

I would conclude that the function is: .$\displaystyle y - 1 \:=\:\frac{-4}{x+1} \quad\Rightarrow\quad\boxed{ y \;=\;\frac{x-3}{x+1}}$

3. Originally Posted by Raoh
$\displaystyle \mathfrak{hello}$
i'm trying to get all the possible informations about a function from its graph and i mean by informations;limits,asymptotes..
i hope someone would help me extract those informations from this example :
Let : $\displaystyle f : \mathbb{R}\to \mathbb{R}\setminus \left \{ -1,3 \right \}$
and here's its graph :
Hello : those informations from this example is:
f(0)=-3
f(3)=0
limf:infini (x--> -1)
limf =1 (x---->infini)

4. thanks guys,but i still don't understand
i'm gonna try to illustrate what i wanted to understand from that graph
$\displaystyle \lim_{x\to +\infty }f(x) = ?$
$\displaystyle \lim_{x\to -\infty }f(x) = ?$
and i don't need just values,i wanna know how
thanks a lot.

5. Originally Posted by Raoh
thanks guys,but i still don't understand
i'm gonna try to illustrate what i wanted to understand from that graph
$\displaystyle \lim_{x\to +\infty }f(x) = ?$
$\displaystyle \lim_{x\to -\infty }f(x) = ?$
and i don't need just values,i wanna know how
thanks a lot.
If $\displaystyle \lim_{x\to\infty}f(x)=L$ then the line $\displaystyle y=L$ is a horizontal asymptote of the graph of $\displaystyle f$.

In your case, it appears from the graph that you have provided, that a good approximation for $\displaystyle L$ is $\displaystyle \frac{1}{2}$

6. Originally Posted by Raoh
thanks guys,but i still don't understand
i'm gonna try to illustrate what i wanted to understand from that graph
$\displaystyle \lim_{x\to +\infty }f(x) = ?$
$\displaystyle \lim_{x\to -\infty }f(x) = ?$
and i don't need just values,i wanna know how
thanks a lot.
Hello : I'have anathor solution
That function is homographic :
1)

2)

3 )
x=-1 is asymtot :
4 ) y=1 is asymtot :

I'have this system :
Subst in f(x) :

conclusion :

7. thanks again for everyone,but i should say that the last post got me a little twisted,because i can't tell from the graph that the function should be written like this $\displaystyle f(x) = \frac{ax+b}{cx+d}$ ,in contrary i like how "Soroban" concluded that the function is indeed $\displaystyle f(x) = \frac{x-3}{x+1}$,but i don't understand his steps , any help ?

8. Originally Posted by Raoh
thanks again for everyone,but i should say that the last post got me a little twisted,because i can't tell from the graph that the function should be written like this $\displaystyle f(x) = \frac{ax+b}{cx+d}$ ,in contrary i like how "Soroban" concluded that the function is indeed $\displaystyle f(x) = \frac{x-3}{x+1}$,but i don't understand his steps , any help ?
Where the lines $\displaystyle y=1$ and $\displaystyle x=-1$ intersect is the center of the hyperbola.
$\displaystyle x+1$ must appear irreducibly in the denominator because the function is undefined at $\displaystyle x=-1$.

He also made use of the intercepts.

9. alright,the domain can tell what should be written in the denominator(i'm not sure if always),what about the numerator ?
thanks.
sorry to bother you guys.