# How do you sketch the general shape of f(x)=x^3-4x^2+7 using end behavior?

Jul 25, 2017

Look below :)

#### Explanation:

Since the first power is odd the general shape of the graph is similar to that of ${x}^{3}$. So the end behavior is ${x}^{3}$. We can also find the y-intercept by replacing all the x-values with zeros:

$f \left(0\right) = {\left(0\right)}^{3} - 4 {\left(0\right)}^{2} + 7$

$y = 7$

All you really need to know to sketch the general shape is the know the end behavior and the y-intercept.

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

This is the graph of $y = {x}^{3} - 4 {x}^{2} + 7$

As you can see the point $\left(0 , 7\right)$ is the y-intercept.

graph{x^3-4x^2+7 [-11.29, 11.21, -3.11, 8.14]}