Sketching a graph.

Question

How would I sketch a graph of $$ f(x)=\frac{x+1}{(x-1)^2} $$ using transformations, not calculus? It should be done with algebraic manipulations, but I haven't had any luck.

Answer 1

You can plot some points. Like at x =1 the function will be undefined (infinite). So for x tending to 1 , the function will increase beyond bounds. At x=0 it will be 1. For x=-1 it will be 0. So you can plot the graph in [-1,1). If you have some predefined interval , you can plot in it.

But , I guess for the entire function you will need basic calculus . Atleast you will have to know in what in what interval is the function increasing , etc.

x=1 will be an asymptote and the graph will fall off on the right & left of it !

Even for x-> infinity , you will have to use L' Hopital ( that's in a way calculus)

Though it's obvious that for x>10 & x< -5 , y ->0

Answer 2

I'm not sure how you would do it with transformations, because I don't know what the parent function would be. But you can make a reasonable sketch without calculus by finding the asymptotes and then testing a few points. Note the following facts:

1. There is a vertical asymptote at $x = 1$ because of the $(x - 1)^2$ in the denominator. So, draw a vertical dotted line at $x = 1$.

2. There is a horizontal asymptote at $y = 0$. Generally, introductory algebra textbooks will simply say that this is true because the leading term in the denominator has a higher power than the leading term in the numerator, so you should be able to justify the horizontal asymptote by this rule. In calculus terms you would say that $\displaystyle \lim_{x \to \infty}f(x) = 0$ and conclude that there is a horizontal asymptote at $y = 0$.

3. The graph approaches $+\infty$ on either side of the vertical asymptote because $f(x) > 0$ if you plug in points just to the left or just to the right of $x = 1$.

4. On the right, the graph approaches the horizontal asymptote from above because plugging in large positive $x$ values makes $f(x) > 0$.

5. On the left, the graph approaches the horizontal asymptote from below because plugging in large negative $x$ values makes $f(x) < 0$.

Using this information you should be able to get a rough idea of what the graph looks like. Draw it as you think it should look and then you can check if your sketch was reasonable by actually plotting the function with a graphing calculator or WolframAlpha.