How do you graph #f(x)=(x^2+2x+1)/x# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Feb 28, 2018

It can be done using the wavy-curve method.

Explanation:

You can't exactly graph it without using a graphing calculator(or by doing all the calculation yourself) but you can come up with a graph which is more or less, the same.

  1. Start by calculating the all the roots of numerator and denominator separately. Here, Numerator has two equal roots(i.e -1) and numerator has root 0.

  2. Now rewrite the equation in this form:
    #y = (x+1)^2/x#

  3. The roots of the denominator is where, you have to draw the vertical asymptotes.

  4. For horizontal asymptotes check #lim_(x -> +-oo) (x+1)^2/x #
    which in this case is #+oo # & # -oo# respectively.

  5. Now we have all the information we need. Mark all the roots on the number line (Only of the numerator), draw vertical asymptotes and the horizontal asymptotes(none in this case)

  6. Now take any value, greater than the biggest root in step one (greatest root in step one is 0, so any value like 1,2,3,etc. would do)
    and check whether our equation gives a positive or a negative result.(In our case it is positive).

  7. Start moving towards the left touching the x- axis at all the roots and moving towards infinity at all vertical asymptotes.

  8. IMPORTANT STEP.. If the term of the root has an even power(2,4,6,etc..) then do NOT cross the x - axis. For example..
    in the equation #(x-2)(x-3)(x-4)^2# , roots are 2,3,4 .. the term which has root 4 is the only term with an even power.. So you cross the x-axis at 2,3 but you bounce off on 4..

This is the wavy curve method..
you should get something like this:
graph{(x+1)^2/x [-10, 10, -5, 5]}