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

Nov 10, 2016

The end behavior of the function is:
As $\rightarrow + \infty$, $f \left(x\right) \rightarrow - \infty$
As $\rightarrow - \infty$, $f \left(x\right) \rightarrow + \infty$

#### Explanation:

If the degree is even, the ends of the graph point in the same direction. If the leading coefficient of an even degree is positive, the ends point up. If the leading coefficient is negative, the ends point down.

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

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

If the degree is odd, the ends of the graph point in opposite directions. If the leading coefficient of an odd degree is positive, the ends of the graph point down on the left and up on the right. If the leading coefficient is negative, the ends of the graph point up on the left and down on the right.

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

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

In the case of $f \left(x\right) = \textcolor{b l u e}{- 1} {x}^{\textcolor{red}{3}} + {x}^{2} + 1$, the degree is $\textcolor{red}{3}$, an odd number, and the leading coefficient is $\textcolor{b l u e}{- 1}$.

So, the end behavior of the function is:
As $\rightarrow + \infty$, $f \left(x\right) \rightarrow - \infty$
As $\rightarrow - \infty$, $f \left(x\right) \rightarrow + \infty$

The graph of the function looks like:

graph{-x^3+x^2+1 [-10, 10, -5, 5]}