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

Jun 25, 2017

The general shape is that of $- {x}^{3}$

Explanation:

Since the first power is odd the general shape of the graph is similar to that of ${x}^{3}$. But we also need to take into account the negative so we say that it behaves like $- {x}^{3}$. We can also find the y-intercept by replacing all the x-values with zeros:

$f \left(0\right) = - {\left(0\right)}^{3} + 2 {\left(0\right)}^{2} + 1$

$y = 1$

If the power is even then it would follow the general shape of ${x}^{2}$.

This is the graph of $- {x}^{3}$
graph{-x^3 [-10, 10, -5, 5]}

This is the graph of $f \left(x\right) = - {x}^{3} + 2 {x}^{2} + 1$
graph{-x^3+2x^2+1 [-10, 10, -5, 5]}