How do you sketch the general shape of #f(x)=x^3-2x^2+1# using end behavior?

1 Answer
Sep 30, 2016

Answer:

#x^3-2x^2+1 \to-\infty# as #x\to-\infty#

#x^3-2x^2+1 \to+\infty# as #x\to+\infty#

Explanation:

You need to know that the end behaviour of a polynomial depends on its degree:

  • if the degree is even, both limits at #\pm\infty# will be #+\infty#
  • if the degree is odd, you'll have the limit according to the direction: if #p(x)# is your polynomial, then #lim_{x\to-\infty}p(x)=-\infty# and #lim_{x\to+\infty}p(x)=+\infty#

This is easy to explain: an even degree means that you surely are the square of something: #x^2# is the square of #x#, #x^4# is the square of #x^2#, and so on. If #n# is even, #x^n# is the square of #x^{n/2}#. And since squares are always positive, the limits can only be #+\infty#.

On the other hand, you can see an odd power as an even power of #x# multiplied one more time by #x#. For example, see #x^7# as #x^6*x#.

We already observed that #x^6# tends to positive infinity in both direction, so at #-\infty# you'll have #x^6*x\to (+\infty)(-\infty)=-\infty#, while at #+\infty# you'll have #x^6*x\to (+\infty)(+\infty)=+\infty#.

The reason for which the leading term is the only relevant one is simple, too: let's analyze your case: we have

#x^3-2x^2+1 = x^3(1-2/x+1/x^3)#

So, if we factor the greatest power of #x#, all the remaining terms will tend to zero as #x# approaches infinity (in both direction), showing that (in this case) #x^3# is the only relevant term to investigate the end behaviour.