How do you find the end behavior of # f(x) = (x+1)^2(x-1) #?

1 Answer
Dec 30, 2015

Answer:

as #xrarr-oo,f(x)rarr-oo#; as #xrarr+oo,f(x)rarr+oo#

Explanation:

The degree of the polynomial is #3#, an odd number.

Since the degree is odd, you know that #f(x)# will approach both #-oo# and #+oo#. In other words, it will end going in different directions—either "up" or "down."

Look at the mother functions:

#f(x)=x^1# (odd)
as #xrarr-oo,f(x)rarr-oo#; as #xrarr+oo,f(x)rarr+oo#
graph{x [-10, 10, -5, 5]}

#f(x)=x^2# (even)
as #xrarr-oo,f(x)rarr+oo#; as #xrarr+oo,f(x)rarr+oo#
graph{x^2 [-10, 10, -5, 5]}

#f(x)=x^3# (odd)
as #xrarr-oo,f(x)rarr-oo#; as #xrarr+oo,f(x)rarr+oo#
graph{x^3 [-10, 10, -5, 5]}

#f(x)=x^4# (even)
as #xrarr-oo,f(x)rarr+oo#; as #xrarr+oo,f(x)rarr+oo#
graph{x^4 [-10, 10, -5, 5]}

Let's examine a cubic: if we change the sign of the first term, what happens?

#f(x)=-x^3#
as #xrarr-oo,f(x)rarr+oo#; as #xrarr+oo,f(x)rarr-oo#
graph{-x^3 [-10, 10, -5, 5]}

However, in our case, we know that the leading term will be positive, so the graph will start "down" and then go "up".

Its end behavior:
as #xrarr-oo,f(x)rarr-oo#; as #xrarr+oo,f(x)rarr+oo#

We can check on a graph:

#f(x)=(x+1)^2(x-1)#
graph{(x+1)^2(x-1) [-10, 10, -5, 5]}