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.
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.
I suppose you know how $\displaystyle f(x)=\frac{1}{x}$ 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 $\displaystyle f(x)=\frac{1}{x+c}$ has a vertical asymptote (the only one) in $\displaystyle x=-c$, so what you have to do is to move (horizontally) $\displaystyle \frac{1}{x}$ so that it has its aymptote there
Remember that they have horizontal asymptotes on both sides
Let's go over the cases.
First $\displaystyle f(x)=\frac{1}{x+2}$
It has a vertical asymptote in $\displaystyle x=-2$ (what would $\displaystyle f(-2)$ be? )
Apart from that, it has the same shape the ones in the other post have.
Now $\displaystyle g(x)=\frac{1}{x-4}$
The vertical asymptote is $\displaystyle x=4$(what would $\displaystyle f(4)$ be? )
Finally $\displaystyle h(x)=\frac{1}{x+1}$
Asymptote $\displaystyle x=-1$
The important thing is that you realise that: $\displaystyle f(x)=g(x+6)$ for example
Just for you to see$\displaystyle f(1)=\frac{1}{1+2}$ and $\displaystyle g(7)=\frac{1}{7-4}$