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

Dec 30, 2015

as $x \rightarrow - \infty , f \left(x\right) \rightarrow - \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow + \infty$

#### Explanation:

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

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

Look at the mother functions:

$f \left(x\right) = {x}^{1}$ (odd)
as $x \rightarrow - \infty , f \left(x\right) \rightarrow - \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow + \infty$
graph{x [-10, 10, -5, 5]}

$f \left(x\right) = {x}^{2}$ (even)
as $x \rightarrow - \infty , f \left(x\right) \rightarrow + \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow + \infty$
graph{x^2 [-10, 10, -5, 5]}

$f \left(x\right) = {x}^{3}$ (odd)
as $x \rightarrow - \infty , f \left(x\right) \rightarrow - \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow + \infty$
graph{x^3 [-10, 10, -5, 5]}

$f \left(x\right) = {x}^{4}$ (even)
as $x \rightarrow - \infty , f \left(x\right) \rightarrow + \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow + \infty$
graph{x^4 [-10, 10, -5, 5]}

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

$f \left(x\right) = - {x}^{3}$
as $x \rightarrow - \infty , f \left(x\right) \rightarrow + \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow - \infty$
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 $x \rightarrow - \infty , f \left(x\right) \rightarrow - \infty$; as $x \rightarrow + \infty , f \left(x\right) \rightarrow + \infty$

We can check on a graph:

$f \left(x\right) = {\left(x + 1\right)}^{2} \left(x - 1\right)$
graph{(x+1)^2(x-1) [-10, 10, -5, 5]}