# Math Help - HELP!!! With graphs (It might be hyberbolic--unsure!)

1. ## HELP!!! With graphs (It might be hyberbolic--unsure!)

I need to know how to sketch graphs of the form y= 1/ x+c:

(i) y= 1/ x+2
(ii) y= 1/ x-4
(iii) y= 1/ x+1

Please help!! I'm clueless. I think they are hyberbolas in the xy=k thing but i cant draw or work it out. Pls show the steps

Thanks.

2. I suppose you know how $f(x)=\frac{1}{x}$ looks like

The cases you posted are equal to this one in shape, mind you they have another position (they are horizontal translations).

The function $f(x)=\frac{1}{x+c}$ has a vertical asymptote (the only one) in $x=-c$, so what you have to do is to move (horizontally) $\frac{1}{x}$ so that it has its aymptote there

Remember that they have horizontal asymptotes on both sides

3. Originally Posted by PaulRS
I suppose you know how $f(x)=\frac{1}{x}$ looks like

The cases you posted are equal to this one in shape, mind you they have another position (they are horizontal translations).

The function $f(x)=\frac{1}{x+c}$ has a vertical asymptote (the only one) in $x=-c$, so what you have to do is to move (horizontally) $\frac{1}{x}$ so that it has its aymptote there

Remember that they have horizontal asymptotes on both sides

But how come the x+3 is situated on top of the x-intercept 2?? and what if it is y= 1/ x-4

the minus sign.....
im so sleepy

4. PLS HELP!!!!!!!
i dunno./.............

im gonna b in big truoouble...omg its so late..........

5. Let's go over the cases.

First $f(x)=\frac{1}{x+2}$

It has a vertical asymptote in $x=-2$ (what would $f(-2)$ be? )

Apart from that, it has the same shape the ones in the other post have.

Now $g(x)=\frac{1}{x-4}$

The vertical asymptote is $x=4$(what would $f(4)$ be? )

Finally $h(x)=\frac{1}{x+1}$

Asymptote $x=-1$

The important thing is that you realise that: $f(x)=g(x+6)$ for example
Just for you to see $f(1)=\frac{1}{1+2}$ and $g(7)=\frac{1}{7-4}$