Drawing graph of function (using basic functions)

[Дьявол]

Draw a graphic of the function using the basic elementary functions.
For example.
Draw $$f(x)=\frac{1}{x-1}-1$$. I know that I will first draw the function 1/x (basic function) then shift the graph of the function 1/x for 1 on right of the x-axis, so I got the graph of the function 1/(x-1). Then shift the whole function for 1 up of the y-axis and I got the graph of f(x).

How will I do the same with $$f(x)=\frac{1}{x^2-1}-1$$.

I tried first with 1/x² then I realized that I can't shift it for 1 on the x-axis? What's the problem? How will I draw the graph?

Thanks in advance.

 

[emilkh]

Draw f(x) = x^2, then shift it down by 1, then draw f(x) = 1/(x^2 -1)

 

[Defennder]

How will I do the same with $$f(x)=\frac{1}{x^2-1}-1$$.

I tried first with 1/x² then I realized that I can't shift it for 1 on the x-axis? What's the problem? How will I draw the graph?

Why can't you shift it? Note that originally the vertical asymptote is at x=0 for 1/x^2. After you shift it, it'll be at $$x = \pm 1$$. Also draw the graph for the interval -1<x<1

 

[DyslexicHobo]

I think in order for you to be able to shift it the same way as 1/(x-1), it would need to be 1/(x-1)^2

Then you could draw the graph of 1/x^2 and shift it by 1

 

[Дьявол]

@emilkh I can't do that, since 1/(x²-1) is not parabola

@Defennder here is the picture.

, you will obviously see that 1/x² is not shifted by the x-axis for 1. It should be 1/(x-1)² so that I can shift it.

@DyslexicHobo, I think also like you. To be possible, it must be 1/(x-1)²

 

[Mentallic]

Is the actual question how does one compare $f(x)\frac{1}{x^2-1}-1$ in terms of the sketch of $f(x)=\frac{1}{x-1}-1$ or how do you just graph the function in general? What you need to look out for etc.

 

[Дьявол]

I need to draw the graphic with shifting. Seems like, the basic graphic would be $$\frac{1}{x^2-1}$$, and then I will shift the graphic for 1 of the y-axis so that I got $$\frac{1}{x^2-1}-1$$

 

[Defennder]

Well the green one looks correct. Now just shift it down by -1 and you're done.

 

[Дьявол]

Yes. So my basic graph would be $$\frac{1}{x^2-1}$$ right?

 

[Defennder]

I'm not sure what you mean by "basic graph", but if you're referring to the graph you start off with before you do the shifting, then I'd say it depends. You can start off from there, or if you're less confident, you can start off with 1/x, then 1/x^2, then 1/(x^2-1) then finally 1/(x^2-1) - 1. Over time when you get more confident and have sufficient practice you can start off with 1/(x^2-1) straight away.

 

[Defennder]

Is the actual question how does one compare $f(x)\frac{1}{x^2-1}-1$ in terms of the sketch of $f(x)=\frac{1}{x-1}-1$ or how do you just graph the function in general? What you need to look out for etc.

The only difference between the first and the second expression is that x has been replaced by x^2. If you think about it, this just means that the graph is symmetrical about the y-axis. Any negative x-value would give a y-value equal to its positive counterpart because of the square term.

To graph such equations you always start off with something known. As above, per what I wrote to Abron, you start off with a few selected graph types for eg. y = x^n where n is odd or when n is even, y = ln(x), y = e^x, y = trigo(x) , where trigo is either sin, cos, tan etc. Then you need to know what replacing x by x-c would do to the graph, what replacing x with x^2 or |x| would do as well.

 

[Дьявол]

Thanks for the replies. I am supposed to solve this problems using the graphics of basic elementary functions. With other words, I should solve it by shifting.

Regards.

 

[Mentallic]

Well ok, if you were to go about graphing this from its 'elementary' graph $$y=\frac{1}{x}$$ then I guess I would go about it something like this:
Extend the hyperbola to $$y=\frac{1}{x^2}$$ by reflecting the negative values of x about the x-axis and if necessary, change the shape of the hyperbola to account for the x squared term.
Extend this again to $$y=\frac{1}{x^2-1}$$ and notice that the denominator can be factorised to $$(x-1)(x+1)$$ so the domain of the function does not exist at 1 and -1. This means you need to shift the previous hyperbola that is asymptotic at x=0 to x=1 for the positive values of x and x=-1 for the negative values. Remember the domain between -1 and 1 though. If you're really unsure what to do in here, I recommend substituting x values slightly less than 1 and more than -1 to find where the range tends to. Also sub x=0 to find where the it cuts the y-axis.
Extend this to $$y=\frac{1}{x^2-1}-1$$ by just shifting everything down 1. i.e. there is an asymptote at y=-1 instead of previously y=0.

 

[Дьявол]

Mentallic thanks for the reply. Ok, now I understand the point.