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

1 Answer
Dec 30, 2015

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]}